= Solution
Use units with $c=1$. For the flat <FRW metric>, the <Friedmann equation> is $H^2=8\pi G\rho/3+\Lambda/3$, where $H=\dot a/a$ is the <Hubble parameter>. The <cosmological continuity equation> for a separately conserved component with constant <equation of state> $P=w\rho$ gives $\rho\propto a^{-3(1+w)}$. Consequently <radiation in cosmology> scales as $a^{-4}$, <pressureless matter> as $a^{-3}$, and <cosmological constant energy> is constant. With $a_0=1$, define the present <critical density> and <cosmological density parameters> by
$$
\rho_{\mathrm{crit},0}=\frac{3H_0^2}{8\pi G},\qquad
\Omega_r=\frac{\rho_{r,0}}{\rho_{\mathrm{crit},0}},\quad
\Omega_m=\frac{\rho_{m,0}}{\rho_{\mathrm{crit},0}},\quad
\Omega_\Lambda=\frac{\Lambda}{3H_0^2}.
$$
The <Hubble constant> $H_0$ measures the expansion rate today. Here $\Omega_r$ includes <photons> and relativistic <neutrinos>, $\Omega_m$ includes <baryons> and <cold dark matter>, and $\Omega_\Lambda$ measures the vacuum contribution. Spatial flatness implies $\Omega_r+\Omega_m+\Omega_\Lambda=1$. Thus
$$
H^2=H_0^2(\Omega_r a^{-4}+\Omega_m a^{-3}+\Omega_\Lambda).
$$
Since <conformal time> satisfies $d\tau=dt/a$, we have $a'=a\dot a=a^2H$. Multiplication by $a^4$ proves
$$
\boxed{a'^2=H_0^2(\Omega_r+\Omega_m a+\Omega_\Lambda a^4).}
$$
For an exam-level estimate, $H_0=100h\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ with $h\simeq0.7$, $\Omega_m\simeq0.3$, $\Omega_\Lambda\simeq0.7$, and the specified <radiation in cosmology> gives $\Omega_r=4.2\times10^{-5}h^{-2}\simeq8.6\times10^{-5}$. These are rounded illustrative flat-universe values; precision <cosmological density parameters> depend on the fitted model. For comparison, the base flat-model fit in https://www.aanda.org/articles/aa/abs/2020/09/aa33910-18/aa33910-18.html[Planck 2018 cosmological parameters] gives $H_0\simeq67.4\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ and $\Omega_m\simeq0.315$.
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