= Solution
For thermal photons, the available thermal length is $\ell_T=\hbar c/(k_BT)$. The number of occupied modes per volume scales as $\ell_T^{-3}$, while their characteristic energy is of order $k_BT=\hbar c/\ell_T$. Thus $n_\gamma\propto\ell_T^{-3}$ and $u_\gamma=\rho_\gamma c^2\propto\hbar c\,\ell_T^{-4}$. More explicitly, the two <photon polarizations> and the <Bose-Einstein distribution> give
$$
n_\gamma=\frac{2\zeta(3)}{\pi^2}\left(\frac{k_BT}{\hbar c}\right)^3,\qquad u_\gamma=\frac{\pi^2}{15}\hbar c\left(\frac{k_BT}{\hbar c}\right)^4,
$$
consistent 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 $Q=k_BT/(\hbar c)$, the photon energy formula gives $Q^4=\Omega_\gamma\rho_cc/(0.66\hbar)$. Therefore
$$
s=\frac{0.24m_p}{\Omega_B\rho_c}\left(\frac{\Omega_\gamma\rho_cc}{0.66\hbar}\right)^{3/4}
=\boxed{\frac{0.24}{0.66^{3/4}}\left(\frac{m_p^4c^3}{\rho_c\hbar^3}\right)^{1/4}\frac{\Omega_\gamma^{3/4}}{\Omega_B}}.
$$
This <photon-to-baryon ratio from density parameters> has leading coefficient $0.328\simeq0.33$, as required. For an illustrative estimate, take the stated $\rho_c=10^{-26}\,\mathrm{kg\,m^{-3}}$, $\Omega_\gamma=5\times10^{-5}$ and $\Omega_B=0.05$. Then $\boxed{s\sim1.4\times10^9}$, 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, $S_\gamma/(k_BN_B)=[4u_\gamma/(3k_BTn_\gamma)]s\simeq3.6s$ with the exact blackbody coefficients. The ratio used in the source is proportional to the radiation entropy per baryon under adiabatic evolution.
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