= Solution
For a relativistic species the supplied relation $P=\epsilon/3$ makes its <entropy density> proportional to $T^3$. The <fermion-to-boson thermal integral ratio> gives the factor $7/8$ for each fermionic degree of freedom, so the electromagnetic plasma immediately before <electron-positron annihilation in cosmology> has
$$
g_{*s,{\rm before}}
=g_\gamma+\frac78(g_{e^-}+g_{e^+})
=2+\frac78(2+2)
=\frac{11}{2}.
$$
After annihilation only the two photon polarizations remain, so $g_{*s,{\rm after}}=2$.
The neutrinos have already undergone <neutrino decoupling>. They therefore receive none of the electron-positron entropy and cool as $T_\nu\propto a^{-1}$. In the still-coupled electromagnetic plasma, <cosmological entropy conservation> gives $g_{*s}T_\gamma^3a^3={\rm constant}$. If $T_d$ and $a_d$ denote the common temperature and scale factor just before annihilation, then at a later scale factor $a$,
$$
2T_\gamma^3a^3=\frac{11}{2}T_d^3a_d^3,
\qquad
T_\nu^3a^3=T_d^3a_d^3.
$$
Dividing gives the <Cosmic neutrino background> temperature ratio
$$
\boxed{\frac{T_\nu}{T_\gamma}=\left(\frac4{11}\right)^{1/3}.}
$$
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