= Solution
Before electron-positron annihilation, the electromagnetic plasma has
$$
g_{*s}=2+\frac78(2+2)=\frac{11}{2},
$$
whereas afterward only the two photon polarizations remain, so $g_{*s}=2$. <Entropy> conservation in the decoupled electromagnetic sector gives
$$
\frac{11}{2}T_{\rm before}^3a^3=2T_\gamma^3a^3.
$$
Neutrinos receive none of this <entropy> and continue cooling as $T_\nu\propto a^{-1}$. Therefore
$$
\boxed{\frac{T_\nu}{T_\gamma}=\left(\frac4{11}\right)^{1/3}.}
$$
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