= Solution
For a relativistic species at zero chemical potential, $p=\epsilon/3$ and $s=(\epsilon+p)/T_i$, so its <entropy density> is proportional to $g_iT_i^3$. The <Bose-Einstein distribution> gives the bosonic normalization, while the fermionic thermal <integral> is $7/8$ of it. Species still in thermal equilibrium have $T_i=T$; freely streaming decoupled species can have another temperature and therefore contribute an extra factor $(T_i/T)^3$ when the total is expressed in <units> of $T^3$. This gives the two terms $N_*$ and $N_{SD}$ and their stated effective-degree-of-freedom weights.
After neutrino decoupling, the <comoving entropy> of the photon-electron-positron plasma is conserved separately from the neutrinos. Before electron-positron annihilation its effective <entropy> count is $2+(7/8)4=11/2$; afterwards it is $2$. Consequently
$$
\frac{11}{2}(aT_{\rm before})^3=2(aT_\gamma)^3,
$$
whereas $aT_\nu$ stays equal to its pre-annihilation value. Hence
$$
\boxed{\frac{T_\nu}{T_\gamma}=\left(\frac4{11}\right)^{1/3}.}
$$
This is the instantaneous-decoupling, negligible residual-electron approximation used in the question. Heating of the still-coupled plasma, not cooling peculiar to neutrinos, creates the temperature difference.
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