Correction to the first printing of third edition: the energy density in the statement of the problem should read

\[
u_{\text {total }}=\left(1+(21 / 8)(4 / 11)^{4 / 3} \right) u_{\gamma} .
\]

After the electron-positron annihilation, the only relativistic species left are the photons and the neutrinos. The factor \(21 / 8=(3)(1)(7 / 8)\) in the energy is because there are three families of neutrinos, the spin degeneracy factor is 1 (all left handed), and \(7 / 8\) is the Fermi-Dirac factor. The factor \((4 / 11)^{4 / 3}\) is due to the lower temperature of the neutrinos compared to the photons; see equation (9.6.4). Following the solution to problem 9.1, we get

\[
t_{0}=\frac{1}{2} \sqrt{\frac{45 \hbar^{3} c^{5}}{8 \pi^{3} G\left(1+\left(\frac{21}{8} \right)\left(\frac{4}{11} \right)^{4 / 3} \right)\left(k T_{0} \right)^{4}}} \simeq 1.79 \mathrm{~s}
\]