= Solution
Before decay, the nonrelativistic $\chi$ density redshifts as $a^{-3}$, while the photon density redshifts as $a^{-4}$. Therefore their ratio grows as $a$. Let a minus sign denote the instant just before decay and define
$$
r_d=\frac{\rho_\chi(t_d^-)}{\rho_\gamma(t_d^-)}
=\frac{a_d}{a_{\mathrm{ref}}}\frac{\rho_\chi(t_{\mathrm{ref}})}{\rho_\gamma(t_{\mathrm{ref}})}.
$$
Using the assumed <radiation domination> law $a\propto t^{1/2}$ gives $r_d=(t_d/t_{\mathrm{ref}})^{1/2}\rho_\chi(t_{\mathrm{ref}})/\rho_\gamma(t_{\mathrm{ref}})$.
Across an instantaneous decay the <scale factor> does not change. Energy conservation and complete thermalization into a zero-chemical-potential photon bath give
$$
\rho_\gamma(t_d^+)=\rho_\gamma(t_d^-)+\rho_\chi(t_d^-)=\rho_\gamma(t_d^-)(1+r_d).
$$
Since <blackbody radiation> has $\rho_\gamma\propto T_\gamma^4$, <instantaneous photon heating by a decaying relic> gives
$$
\boxed{\frac{T_\gamma^+}{T_\gamma^-}=(1+r_d)^{1/4}.}
$$
The already decoupled <Cosmic neutrino background> is not heated. Dividing its unchanged temperature by the new photon temperature therefore yields
$$
\boxed{\frac{T_\nu}{T_\gamma}=\left(\frac4{11}\right)^{1/3}
\left[1+\left(\frac{t_d}{t_{\mathrm{ref}}}\right)^{1/2}
\frac{\rho_\chi(t_{\mathrm{ref}})}{\rho_\gamma(t_{\mathrm{ref}})}\right]^{-1/4}.}
$$
The relation written with $a_d/a_{\mathrm{ref}}$ is independent of the expansion approximation. Replacing it by the time ratio uses the radiation-dominated background assumed in the question; an appreciable matter contribution would require solving for that background instead. This photon entropy injection is not adiabatic, so the pre-decay photon entropy need not be conserved across the decay.
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