= Solution
Take an unresolved spherical planet at <Euclidean distance> $d\gg R$, with uniform surface <temperature> $T$, stationary isotropic emission, and no intervening extinction. Isothermality does not by itself imply <blackbody> radiation; first derive the geometric result for a uniform emergent <specific intensity> $I_\lambda$.
The observer sees the projected disc with apparent <solid angle> $\Omega_p\simeq\pi R^2/d^2$. Integrating its <specific intensity> over that angle gives the observed spectral <radiative flux>
$$
f_\lambda=\int_{\rm disc}I_\lambda\cos\theta_{\rm obs}\,d\Omega
\simeq\pi I_\lambda\frac{R^2}{d^2},
$$
where $\theta_{\rm obs}$ is the small angle to the detector normal. For a <blackbody>, $I_\lambda=B_\lambda(T)$, and the <Stefan–Boltzmann law> gives $\int_0^\infty\pi B_\lambda(T)\,d\lambda=\sigma_{\rm SB}T^4$. Hence
$$
\boxed{f_\lambda=\pi B_\lambda(T)(R/d)^2,\qquad
f=\sigma_{\rm SB}T^4(R/d)^2=\frac{L}{4\pi d^2}.}
$$
Equivalently, the surface <radiative flux> is $\sigma_{\rm SB}T^4$ and the <luminosity of a spherical blackbody> is $L=4\pi R^2\sigma_{\rm SB}T^4$, leading to the same result by inverse-square dilution. A uniform wavelength-dependent <emissivity> $\epsilon_\lambda$ multiplies the spectral formula; an atmospheric spectrum can require a full angular <radiative transfer> integral rather than a single <Planck function>.
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