Here's my approach to solving this problem. You provide some of the initial steps, but I'll still go through them just for completeness.
Orbital Distance
We're told the period is $T=5.5\:\mathrm{years}$. This means we can immediately calculate the orbital distance (or more precisely, the semi-major axis, $a$). Since we're talking about a comet orbiting the Sun, we can simply use:
$$T^2 = a^3$$
where $T$ is the orbital period in units of $\mathrm{years}$ and $a$ is the semi-major axis in units of $\mathrm{AU}$. I find that $a = 3.116\:\mathrm{AU} = 4.66\times 10^{11}\:\mathrm{m}$. Good so far.
Volume of Evaporated Shell
We're told that the radius of the comet decreases by $\Delta R = 20\:\mathrm{cm}$. We can assume the comet is perfectly spherical and calculate the volume of the evaporated shell which should depend on both $R$, the radius of the comet, and $\Delta R$. This will be necessary since we need to know the total amount of ice that was evaporated. The volume of this shell is given by
$$V = \frac{4}{3}\pi(R^3-(R-\Delta R)^3)$$
$$V = \frac{4}{3}\pi(R^3-R^3+3R^2\Delta R-3R\Delta R^2 + \Delta R^3)$$
$$V = \frac{4}{3}\pi(3R^2\Delta R-3R\Delta R^2)$$
$$V = 4\pi R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$
Note that I've made a specific choice here. I've dropped the $\Delta R^3$ term from the second line. The reason being that it is a third order term and $\Delta R^3 \ll R\Delta R^2$. You could argue that I could drop the second order term, $R\Delta R^2$ since $R\Delta R^2 \ll R^2\Delta R$, but I'm choosing to keep this second order term around so we wind up with an $R$ to solve for in the final answer.
Power Input
The next step is to find the total energy input per second to this comet, e.g., the power. You've basically already defined this part. The flux on this comet is defined as
$$F = \frac{L_\odot}{4\pi a^2}$$
The power input is simply the flux times the area of the comet.
$$P_{\mathrm{in}} = FA = \frac{L_\odot}{4\pi a^2} \pi R^2 = \frac{1}{4}L_\odot \frac{R^2}{a^2}$$
Note that this assumes the comet is in a circular orbit and thus always at the orbital radius of $a$. If the comet's orbit had any sort of eccentricity, then $F$ would be a function of radius and you'd have a much harder time.
Evaporation Energy
Now we need to calculate the total energy necessary to evaporate the evaporated shell of volume $V$ from above. In order to evaporate a solid ice, you have to go through four stages of heating. First, you raise the ice's temperature to the melting point. The energy input for this is defined by the specific heat capacity for ice, $c_{\mathrm{ice}}$. Then you add energy to convert the ice into water. The energy input for this is defined by the latent heat of fusion, $\ell_f$. Now you can raise the temperature of the water until it reaches the next stage. This is defined by the specific heat of water, $c_{\mathrm{water}}$. Finally, you add energy to convert the water to gas, defined by the latent heat of vaporization, $\ell_v$.
All of these can be put together into a single equation.
$$E_{\mathrm{evap}} = c_{\mathrm{ice}}(m_{\mathrm{shell}}\Delta T_1) + \ell_fm_{\mathrm{shell}}+c_{\mathrm{water}}(m_{\mathrm{shell}}\Delta T_2) + \ell_vm_{\mathrm{shell}}$$
$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)m_{\mathrm{shell}}$$
Each one of the terms in this equation accounts for the energy input of one of the stages I described above. Note that $\Delta T_1$ is the temperature change from the initial temperature to the melting point ($273.15\:\mathrm{K}$). A reasonable temperature change might be $73.15\:\mathrm{K}$ (assuming an initial temperature of $200\:\mathrm{K}$), based on the temperature of comet 67P as determined by Rosetta. The $\Delta T_2$ is the temperature change from the melting point to the boiling point which is necessarily $100\:\mathrm{K}$.
You can look in a table somewhere and find that $c_{\mathrm{ice}} = 2.108\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$ and $c_{\mathrm{water}} = 4.187\:\mathrm{kJ\:kg^{-1}\:K^{-1}}$.
Lastly, you need to define the mass of the evaporated shell, $m_{\mathrm{shell}}$. This is simply the volume already determined, multiplied by the density of ice/water. Technically those densities will be different, but they're close enough that we can just use $\rho_{shell} = 1000\:\mathrm{kg\:m^3}$. So ultimately we have:
$$E_{\mathrm{evap}} = (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)V \rho_{\mathrm{shell}}$$
For simplicity's sake, I'm going to define
$$\eta \equiv (c_{\mathrm{ice}}\Delta T_1 + \ell_f +c_{\mathrm{water}}\Delta T_2 + \ell_v)$$
so that
$$E_{\mathrm{evap}} = 4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)$$
Putting it all together
We now know the total energy necessary to evaporate the comet's $20\:\mathrm{cm}$ shell as well as the energy input per second. We know that it receives this energy input per second for one orbital period of $5.5\:\mathrm{years}$ which means we can say:
$$T = \frac{E_{\mathrm{evap}}}{P_{\mathrm{in}}}$$
$$T = \frac{4\pi \eta \rho_{\mathrm{shell}} R^2\Delta R\left(1-\frac{\Delta R}{R}\right)}{\frac{1}{4}L_\odot \frac{R^2}{a^2}}$$
$$T = 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot} \left(1-\frac{\Delta R}{R}\right)$$
Again, for simplicity, I'll define
$$\xi \equiv 8\pi \eta \rho_{\mathrm{shell}} \Delta R \frac{a^2}{L_\odot}$$
so that
$$T = \xi \left(1-\frac{\Delta R}{R}\right)$$
It should be pretty easy to see now that
$$\boxed{R = \frac{\Delta R}{1-T/\xi}}$$
The rest is just plugging everything in.
