= Solution
For a relativistic equilibrium species with temperature $T_i$, the entropy density is $s_i=(\rho_i+P_i)/T_i=4\rho_i/(3T_i)$. In units $k_B=\hbar=c=1$, bosons have $\rho_i=\pi^2g_iT_i^4/30$, while fermions have $7/8$ times this expression. Relative to a reference bath temperature $T$, define the <effective number of relativistic degrees of freedom> for entropy by
$$
\boxed{g_{*s}=\sum_{\mathrm{bosons}}g_i\left(\frac{T_i}{T}\right)^3
+\frac78\sum_{\mathrm{fermions}}g_i\left(\frac{T_i}{T}\right)^3,\qquad
s=\frac{2\pi^2}{45}g_{*s}T^3.}
$$
The sums here concern relativistic species, or their effective entropy contributions. Since a physical comoving patch has volume proportional to $a^3$, <cosmological entropy conservation> gives
$$
\boxed{g_{*s}a^3T^3=\text{constant}}
$$
for an isolated adiabatic bath. After <thermal decoupling in cosmology>, one must apply this law separately to sectors that no longer exchange entropy.
Before <electron-positron annihilation in cosmology>, the electromagnetic bath contains two photon polarizations and four electron/positron spin states. After <neutrino decoupling>, its entropy degrees of freedom are initially
$$
g_{*s,\mathrm{before}}=2+\frac78(4)=\frac{11}{2},
$$
and after annihilation only the two photon polarizations remain. Electromagnetic entropy conservation therefore gives
$$
2(aT_\gamma)_{\mathrm{after}}^3=\frac{11}{2}(aT_\gamma)_{\mathrm{before}}^3.
$$
The decoupled <Cosmic neutrino background> receives none of this entropy; its distribution temperature obeys $aT_\nu=\text{constant}$, and $T_\nu=T_\gamma$ before annihilation. Thus
$$
\boxed{\frac{T_\nu}{T_\gamma}=\left(\frac4{11}\right)^{1/3}.}
$$
This is the <temperature of a decoupled relativistic relic> in the instantaneous-decoupling approximation. The ratio remains the same afterwards in the absence of additional photon heating. At late times $T_\nu$ denotes the redshifted relic distribution temperature even when neutrino masses become relevant; it need not be a kinetic-equilibrium temperature.
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