= Solution
Evaluate <radiation pressure> on the gas-supported <hydrostatic equilibrium> solution. At the centre,
$$
\frac{P_{{\rm rad},c}}{P_{{\rm gas},c}}=\frac{aT_c^4}{3P_c}
=\boxed{\frac{125\pi aG^3}{15552\mathcal R^4}\,M^2}.
$$
The factors $R^{-4}$ cancel between $T_c^4$ and $P_c$, so at fixed <mean molecular weight> the ratio depends only on mass. Equivalently it is $(M/M_{\rm eq})^2$, with
$$
\boxed{M_{\rm eq}=\sqrt{\frac{15552}{125\pi a}}\,\frac{\mathcal R^2}{G^{3/2}}=\sqrt{\frac{15552}{125\pi a}}\,\frac{k^2}{G^{3/2}\mu^2H^2}.}
$$
For $H\simeq1.6605\times10^{-27}\,\mathrm{kg}$ this is approximately $14.6\mu^{-2}M_\odot$, or $40.5M_\odot$ if $\mu=0.6$. If $H$ denotes hydrogen-atom mass instead, use the corresponding <mean molecular weight> convention consistently; the coefficient is about $14.4\mu^{-2}M_\odot$. \b[Equality marks the breakdown of the gas-pressure approximation.] It is a diagnostic based on the gas-only model, not an exact mixed-pressure stellar solution.
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