= Solution
At full phase, reflected-light <contrast> is
$$
\boxed{\frac{F_{p,\lambda}^{\rm refl}}{F_{\star,\lambda}}=A_{g,\lambda}\left(\frac{R_p}{d}\right)^2,}
$$
where $A_g$ is the <geometric albedo>, and other phases multiply this by a <phase function> normalized to one at full phase. Thermal <contrast> for uniform blackbody emitters is
$$
\boxed{\frac{F_{p,\lambda}^{\rm th}}{F_{\star,\lambda}}=\left(\frac{R_p}{R_\star}\right)^2\frac{B_\lambda(T_p)}{B_\lambda(T_{\rm eff})}.}
$$
The corresponding bolometric thermal ratio is $(R_p/R_\star)^2(T_p/T_{\rm eff})^4$. If zero internal heating and equilibrium are assumed, energy balance also gives a bolometric reflected ratio $\alpha(R_p/d)^2/4=(\alpha\epsilon/(1-\alpha))(R_p/R_\star)^2(T_p/T_{\rm eff})^4$ for the luminosity ratio; it is not a directional full-phase <contrast>.
Earth's thermal emission peaks around $10\,\mu\mathrm m$, and solar emission around $0.5\,\mu\mathrm m$. At $0.5\,\mu\mathrm m$ the thermal <contrast> for $T_p=288\,\mathrm K$ is negligible, about $5\times10^{-46}$; reflected light dominates. Taking illustrative visible $A_g\simeq0.3$ gives $\boxed{C_{0.5}\sim5\times10^{-10}}$. A <Bond albedo> does not uniquely determine a spectral <geometric albedo>; a Lambert sphere with Bond albedo $0.3$ would instead have $A_g=0.2$, giving $3.6\times10^{-10}$. At $20\,\mu\mathrm m$, the <Planck function> gives $\boxed{C_{20}\simeq1.0\times10^{-6}}$ for $288\,\mathrm K$; using $255\,\mathrm K$ gives about $7\times10^{-7}$. These are order estimates, since the terrestrial <spectrum> is not a uniform blackbody.
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