= Solution
The fluid conservation equation is $\dot\rho+3H(\rho+P/c^2)=0$. With the radiation equation of state, it becomes $\dot\rho+4H\rho=0$, giving $\rho a^4=\rho_0$. Define the <critical density> $\rho_{\rm crit}=3H^2/(8\pi G)$ and <cosmological density parameter> $\Omega=\rho/\rho_{\rm crit}$. Since $a_0=1$,
$$
\boxed{\rho=\frac{3H_0^2\Omega_0}{8\pi G a^4}.}
$$
Evaluating the <Friedmann equation> today gives $H_0^2+kc^2=H_0^2\Omega_0$, hence
$$
\boxed{kc^2=H_0^2(\Omega_0-1),\qquad
H^2=H_0^2\Omega_0\left(a^{-4}-\beta a^{-2}\right),
\quad\beta=\frac{\Omega_0-1}{\Omega_0}.}
$$
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