= Solution
{c}
Use one <spectral resolution element>, $\Delta\nu=0.5\,\mathrm{cm}^{-1}$, and assume comparable angular resolution horizontally and vertically, so the pixel <solid angle> is $\Omega\simeq\theta^2=6.67\times10^{-6}\,\mathrm{sr}$. The question specifies only the vertical angle; without an assumption about horizontal field of view the diameter is not unique. Interpret throughput $\eta=0.5$ as dimensionless transmission. With primary collecting area $A_p=\pi d^2/4$, the power in that element is
$$
P_s=\eta B_\nu\Delta\nu\,A_p\Omega
=1.30\times10^{-7}A_p\,\mathrm W,
$$
when $A_p$ is in square metres. There is no extra factor of $\pi$ multiplying the supplied <spectral radiance>.
For an integration time $t=1$ s, the order-of-magnitude <effective noise bandwidth> is $\Delta f\simeq1/t=1$ Hz. The root-mean-square <noise-equivalent power> is then $P_n={\rm NEP}\sqrt{\Delta f}=5\times10^{-11}$ W.<NEP – Noise Equivalent Power by Thorlabs>[p. 1, NEP Definition; p. 2, equations (2)–(3): minimum power and averaging time]{ref} A <signal-to-noise ratio> of 10 requires $P_s=5\times10^{-10}$ W, and therefore
$$
\boxed{A_p\simeq3.84\times10^{-3}\,\mathrm{m^2},\qquad d\simeq7.0\,\mathrm{cm}.}
$$
An ideal rectangular average of a one-sided white <noise> spectral density has $\Delta f=1/(2t)$, giving $d\simeq5.9$ cm instead. A circular field with diameter $\theta$ changes the diameter by another order-unity factor. These convention choices all require an aperture substantially larger than the $9.5$ mm <diffraction> estimate, so signal collection is the stronger constraint. The estimate treats the quoted NEP as applicable to the reconstructed channel; a detailed <Fourier transform spectroscopy> design would also specify scan weighting, background and how interferogram noise propagates into that channel.
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