= Solution
As the plasma cools through electron-positron annihilation, the pairs transfer their <entropy> to the coupled <photons>. <Neutrinos> have largely decoupled and do not share that heating in the instantaneous-decoupling approximation. Before annihilation the electromagnetic <entropy> count is
$$
g_{*s,\rm before}=2+\frac78(2+2)=\frac{11}{2},
$$
while after it $g_{*s,\rm after}=2$. Conservation of electromagnetic <comoving entropy>, $g_{*s}T_\gamma^3a^3=\text{constant}$, gives the relative <photon> heating factor
$$
\boxed{\frac{(aT_\gamma)_{\rm after}}{(aT_\gamma)_{\rm before}}
=\left(\frac{11}{4}\right)^{1/3}\simeq1.401.}
$$
Equivalently $T_\gamma/T_\nu=(11/4)^{1/3}$ afterward. The exact $11/4$ count compares an asymptotic relativistic pair bath with the pair-free bath; at the supplied <temperature> some finite-electron-mass <entropy> loss has already begun.
The word “rises” requires care. The <photons> become hotter than an otherwise freely redshifting bath at the same <scale factor>; their absolute <temperature> need not rise during the expansion. In the smooth equilibrium process,
$$
\boxed{\frac{d\ln T_\gamma}{d\ln a}
=-\frac{3}{3+d\ln g_{*s}/d\ln T_\gamma}<0,}
$$
since the pair contribution grows with <temperature>. This <monotonic photon cooling during pair annihilation> is slower than $a^{-1}$ while $aT_\gamma$ increases. The factor $1.401$ compares the <photon> <temperature> with the freely redshifting <neutrino> bath, not two consecutive absolute temperatures along the expanding history.
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