Approximating density of hydrogen in [observable] universe

Question

Let universe be completely made from hydrogen. And also we have redshift $z= 6$. with Hubble constant $H_{0} = 2.1941747572815535\times 10^{-18}\:\mathrm{s}^{-1}$. We also know that density of the universe $\rho =\frac{3H^2}{8\pi G}$ ($G= 6.67\times 10^{-11}\:\mathrm{m^3\ kg^{-1}\ s^{-2}}$).

Is the following approach right to find the number density of hydrogen in universe?

First we calculate density $\rho =\frac{3H^2}{8\pi G}$, then we have $1+z = \sqrt{\frac{1+v/c}{1-v/c}}$. From that we get $v = H\times R$ and from this we calculate $R$, the radius of observable universe. With that, we calculate volume of universe $V = (4\pi/3)R^3$ and we get Mass $M=\rho V$. Then we calculate number of Hydrogen atoms with $N = M/m_{\mathrm{hydrogen}}$. Now we have number density of hydrogen $\rho = N/V$.

Is this solution right? Or we must solve it another way?

_{The source of question is National Olympiad of Iran which is not available in English so the above is my translation for the question.}

Answer 1

So to estimate the answer, you need to convert the mass density to number density by $n_H = \rho_H / m_H$. After getting the current number density you need to redshift it to finite $z$, here $z=6$. Keep in mind the density of hydrogen evolves $\sim (1+z)^3$ due to the expansion of the universe.

That said, the assumption 'let universe be completely made from hydrogen' renders this question highly unrealistic, hence I agree with @James K that this is not real world astronomy. In the real world, only about 5% of the total energy is made of baryons, among which Hydrogen abundance is $\sim 75\%$. Also, at redshift $z\sim 6$, star formation etc. lowers the neutral hydrogen gas density, c.f. recent observations here and here.