 Solution

For thermal photons, the available thermal length is ${\ell }_T=\hslash c/(k_{B}T)$. The number of occupied modes per volume scales as ${\ell }_T^{-3}$, while their characteristic energy is of order $k_{B}T=\hslash c/{\ell }_T$. Thus $n_{\gamma }\propto {\ell }_T^{-3}$ and $u_{\gamma }={\rho }_{\gamma }{c}^2\propto \hslash c{\ell }_T^{-4}$. More explicitly, the two photon polarizations and the Bose-Einstein distribution giveconsistent with the rounded coefficients supplied in the source.

Use its definition $s=n_{\gamma }/n_B$, with $n_B={\rho }_B/m_p={\Omega }_B{\rho }_c/m_p$ and ${\rho }_c$ the mass-equivalent cosmological critical density. Writing $Q=k_{B}T/(\hslash c)$, the photon energy formula gives $Q^4={\Omega }_{\gamma }{\rho }_{c}c/(0.66\hslash )$. ThereforeThis photon-to-baryon ratio from density parameters has leading coefficient $0.328\simeq 0.33$, as required. For an illustrative estimate, take the stated ${\rho }_c=10^{-26}\mathrm{kg}{\mathrm{m}}^{-3}$, ${\Omega }_{\gamma }=5\times 10^{-5}$ and ${\Omega }_B=0.05$. Then $\boxed{s\sim 1.4\times 10^9}$, i.e. of order a billion photons per baryon. The source alone supplies no numerical density parameters, so an estimate must state such additional inputs.

Strictly, this $s$ is the photon-to-baryon ratio, rather than the thermodynamic entropy per baryon itself. For a thermal photon gas, $S_{\gamma }/(k_B{N}_B)=[4u_{\gamma }/(3k_{B}T{n}_{\gamma })]s\simeq 3.6s$ with the exact blackbody coefficients. The ratio used in the source is proportional to the radiation entropy per baryon under adiabatic evolution.