= Solution
Before <electron-positron annihilation in cosmology>, the pair has two spin states, just as <photons> do, and the fermionic number integral gives $n_{\nu+\bar\nu}/n_\gamma=3/4$ at equal temperature. The factor follows by expanding $(e^x+1)^{-1}$ into an alternating exponential series: its $x^2$ integral is $3/4$ of the bosonic one.
After <neutrino decoupling>, <cosmological entropy conservation> in the photon-electron plasma changes its relativistic entropy count from $2+(7/8)4=11/2$ to $2$. The decoupled <neutrinos> do not share this heating. Hence
$$
\left(\frac{T_\nu}{T_\gamma}\right)^3=\frac4{11},\qquad
\boxed{n_{\nu+\bar\nu,0}=\frac34\frac4{11}n_{\gamma0}
=\frac3{11}(410)\simeq111.8\,\mathrm{cm}^{-3}.}
$$
These counts include both the new <neutrino> and its <antineutrino>; another factor of two would double count them. This is the <number density of a relativistic neutrino relative to photons>, evaluated after photon heating.
They are nonrelativistic today, so the <relic-neutrino energy density> is dominated by rest mass:
$$
\rho_{\nu0}\simeq (10^3\,\mathrm{eV})(111.8\,\mathrm{cm}^{-3})
=1.118\times10^5\,\mathrm{eV}\,\mathrm{cm}^{-3},\qquad
\boxed{\Omega_{\nu0}\simeq22.4.}
$$
The number follows from the supplied <critical density>, without a Hubble-unit convention. It is conditional on the thermal history stated above; a sterile, colder or nonthermally produced state need not have this abundance.
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