 Solution

For a separately conserved homogeneous component, energy conservation gives $d(\rho c^2{a}^3)=-pd(a^3)$, equivalently $\dot{\rho }+3H(\rho +p/c^2)=0$. Dust has negligible pressure, while isotropic radiation has $p_{\gamma }={\rho }_{\gamma }{c}^2/3$. HenceThe matter factor is volume dilution; photons additionally lose energy by cosmological redshift. Nonrelativistic baryons and cold dark matter both contribute to the matter term, so ${\Omega }_M={\Omega }_B+{\Omega }_C$. At equality ${\rho }_{M0}{a}_{\mathrm{eq}}^{-3}={\rho }_{\gamma 0}{a}_{\mathrm{eq}}^{-4}$, givingThis is equality for the source's matter-plus-photon model. Additional relativistic species would contribute to total radiation and shift the equality value. Although the source calls the $\rho$'s energy densities, its Friedmann equation and later ${\rho }_{\gamma }{c}^2$ show that their consistent SI meaning is mass-equivalent density.