Answer
The critical density is $~~5.7~~$ hydrogen atoms per $m^3$
Work Step by Step
We can find evaluate the density $\rho$ numerically:
$\rho = \frac{3~H^2}{8~\pi~G}$
$\rho = \frac{(3)~(0.0218~m/s~ly)^2}{(8~\pi)~(6.67\times 10^{-11}~N~m^2/kg^2)}$
$\rho = \frac{(3)~(0.0218~m/s~ly)^2(\frac{1~ly}{9.461\times 10^{15}~m})^2}{(8~\pi)~(6.67\times 10^{-11}~N~m^2/kg^2)}$
$\rho = 9.50\times 10^{-27}~kg/m^3$
$\rho = (9.50\times 10^{-27}~kg/m^3)(\frac{1~H~atom}{1.67\times 10^{-27}~kg})$
$\rho = 5.7~H~atom/m^3$
The critical density is $~~5.7~~$ hydrogen atoms per $m^3$
