 Solution

For a uniformly illuminated single slit of width $w$, the far-field amplitude is proportional to $\sin u/u$, with $u=\pi w\sin \vartheta /\lambda$. Its first zeros occur at $\sin \vartheta =\pm \lambda /w$. If the quoted peak width is the full width between those zeros, the small-angle geometry givesThe Fraunhofer diffraction approximation is appropriate since $w^2/(\lambda L)\ll 1$. If the supplied width instead meant a half-width or full width at half maximum, its numerical conversion factor would differ; the displayed answer states the first-zero convention explicitly.