= 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 gives
$$
\Delta y\simeq\frac{2L\lambda}{w},\qquad\boxed{w\simeq\frac{2L\lambda}{\Delta y}\simeq2.3\times10^{-8}\,\mathrm m.}
$$
The <Fraunhofer diffraction> approximation is appropriate since $w^2/(\lambda L)\ll1$. 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.
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