You could use the Bondi-Hoyle accretion rate: $$\dot{M} = \pi \rho \frac{G^2 M^2}{c_s^3}\ ,$$ where $\rho$ is the density of the solar material and $c_s$ is its sound speed.

The biggest values will occur at the centre because the density is greatest at the centre and increases more rapidly than the cube of the sound speed. Using rough values of say, $c_s \sim 500$ km/s and a density of $10^5$ kg/m$^3$, we get, for $M=6\times 10^{24}$ kg, $$ \dot{M} = 4\times 10^{17} \left(\frac{\rho}{10^3\ {\rm kg\ m}^{-3}}\right)\left( \frac{c_s}{300\ {\rm km\ s}^{-1}}\right)^{-3}\ {\rm kg\ s}^{-1} $$

The black hole would be travelling at around 600 km/s, so would traverse the central 10% of the Sun in around 100s (the accretion rate gets orders of magnitude lower in the outer parts of the Sun, so can be neglected), so might increase its mass by $4\times 10^{19}$ kg (or 0.0007%). This would be insufficient to slow it appreciably and it would shoot out of the other side and disappear into space again.

All this assume that the accretion can occur at the Bondi-Hoyle rate. This i svery unlikely for two reasons - the first is turbulence which reduces the accretion efficiency. The second is radiation pressure. There will be some sort of super-heated region around the black hole and the radiation from this region will exert an outward pressure on the surrounding gas. The limiting Eddington accretion rate, where the radiation pressure prevents a higher rate of infall is $$ \dot{M} \leq \frac{4 \pi G M m_p}{\epsilon c \sigma_T}\ ,$$ where $m_p$ is the proton mass and $\sigma_T = 6.6\times 10^{-29}$ m$^2$ is the Thomson scattering cross-section for free electrons in the solar plasma. The $\epsilon$ parameter is the radiative efficiency of the infalling gas and is usually assumed to be of order 0.1. Taken together, this gives a maximum accretion rate of $$ \dot{M} = 4 \times 10^9 \left(\frac{\epsilon}{0.1}\right)^{-1}\ {\rm kg\ s}^{-1}$$

This is way lower than the Bondi-Hoyle rate. I conclude that the black hole would shoot through the Sun and leave on the other side, having hardly accreted any mass (as a fraction of its own mass).

The answer is that there is a limited rate at which matter that can be crammed into a black hole of this size. That rate is small enough that the black hole will traverse through the Sun hardly picking up any mass (when expressed as a fraction of its original mass).

Details:

You could use the Bondi-Hoyle accretion rate: $$\dot{M} = \frac{dM}{dt} = \pi \rho \frac{G^2 M^2}{c_s^3}\ ,$$ where $\rho$ is the density of the solar material and $c_s$ is its sound speed.

The biggest values will occur at the centre because the density is greatest at the centre and increases more rapidly than the cube of the sound speed. Using rough values of say, $c_s \sim 500$ km/s and a density of $10^5$ kg/m$^3$, we get, for $M=6\times 10^{25}$ kg, $$ \dot{M} = 4\times 10^{19} \left(\frac{\rho}{10^3\ {\rm kg\ m}^{-3}}\right)\left( \frac{c_s}{300\ {\rm km\ s}^{-1}}\right)^{-3}\ {\rm kg\ s}^{-1} $$

The black hole would be travelling at around 600 km/s, so would traverse the central 10% of the Sun in around 100s (the accretion rate gets orders of magnitude lower in the outer parts of the Sun, so can be neglected), so might increase its mass by $4\times 10^{21}$ kg (or 0.007%). This would be insufficient to slow it appreciably and it would shoot out of the other side and disappear into space again. Note that a larger black hole would pick up more mass as a fraction of its original mass because $\dot{M} \propto M^2$.

All this assumes that the accretion can occur at the Bondi-Hoyle rate. This is very unlikely for two reasons - the first is turbulence which reduces the accretion efficiency. The second is radiation pressure. There will be some sort of super-heated region immediately around the black hole (just imagine trying to compress $4 \times 10^{19}$ kg of material into a 20 cm diameter hole every, the event horizon diameter for a 10 Earth-mass black hole, every second) and the radiation from this region will exert an outward pressure on the surrounding gas. The limiting Eddington accretion rate, where the radiation pressure from the hot gas immediately around the black hole, exerts enough pressure to halt further infalling gas, is given by $$ \dot{M} \leq \frac{4 \pi G M m_p}{\epsilon c \sigma_T}\ ,$$ where $m_p$ is the proton mass and $\sigma_T = 6.6\times 10^{-29}$ m$^2$ is the Thomson scattering cross-section for free electrons in the solar plasma (i.e. the effective area an electron presents to radiation). The $\epsilon$ parameter is the radiative efficiency of the infalling gas (i.e. what fraction of its kinetic energy is radiated away) and is usually assumed to be of order 0.1. Taken together, this gives a maximum accretion rate (for a ten Earth-mass black hole) of $$ \dot{M} = 4 \times 10^{10} \left(\frac{\epsilon}{0.1}\right)^{-1}\ {\rm kg\ s}^{-1}$$

Whether the Eddington limit scales down to micro black holes and dense environments like this, I am unsure. But it seems clear that the accretion rate will be way lower than the Bondi-Hoyle rate. I conclude that the Bondi-Hoyle accretion rate is probably a vast over-estimate, and that the black hole would shoot through the Sun and leave on the other side, having hardly accreted any mass (as a fraction of its own mass).