 Solution

Interpret the stated detector width $d=20\mu \mathrm{m}$ as one square spatial sample. At the point directly below the spacecraft, the small-angle approximation gives pixel angular width $d/f\simeq \Delta x/h$. ThusThere is a definition error in the source: the usual focal ratio, or f-number, divides focal length by the entrance-aperture diameter $D$, whereas the printed parenthesis divides by detector width $d$. Reporting both avoids silently changing the question:The actual optical design is therefore approximately f/1.36, a fast optical telescope, not an optically slow f/34000 instrument. Such a focal ratio is demanding but possible in principle: optical aberrations, obscuration, field coverage, stray light and alignment need careful design. The single-sample angular width is $2.94\times 10^{-5}\mathrm{rad}$. Spatial sampling alone does not guarantee true resolving power; diffraction, pointing and signal-to-noise ratio also matter. At large viewing angles, the same angular pixel covers a larger, distorted surface footprint. If the detector is literally one element, scanning is additionally needed to form a map.