= Solution
The <planetary equilibrium temperature> is the effective planetary <temperature> at which absorbed starlight balances emitted thermal radiation, neglecting internal power. With uniform reradiation over the full surface,
$$
(1-\alpha)\pi R_p^2\sigma T_{\rm eff}^4\frac{R_\star^2}{d^2}=4\pi R_p^2\epsilon\sigma T_{\rm eq}^4.
$$
Cancelling areas gives
$$
\boxed{T_{\rm eq}=T_{\rm eff}\left(\frac{R_\star}{2d}\right)^{1/2}\left(\frac{1-\alpha}{\epsilon}\right)^{1/4}.}
$$
Using the supplied <Bond albedos> and taking the thermal <emissivity> $\epsilon=1$ yields $\boxed{T_{\rm eq,\oplus}\simeq255\,\mathrm K,\quad T_{\rm eq,J}\simeq103\,\mathrm K}$. “Perfect black bodies” must here mean perfect infrared emitters, since a perfect absorber at all wavelengths would instead have $\alpha=0$, inconsistent with retaining the supplied reflected fractions. If both albedos were set to zero, the corresponding values would be about $279$ and $122\,\mathrm K$.
Earth's mean surface <temperature> of about $288\,\mathrm K$ exceeds its emission <temperature> mainly because of the <greenhouse effect>. Jupiter also radiates substantial internal <energy> from cooling and contraction, including interior differentiation processes, so absorbed sunlight alone underpredicts its emission. Jupiter has no solid surface; the quoted $150\,\mathrm K$ is a representative atmospheric level and is not identical to an effective radiating <temperature>.
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