Answer
$1.7\mu m$
Work Step by Step
We can find the required width of the slit as follows:
$W=\frac{\lambda}{sin\theta}$
We plug in the known values to obtain:
$W=\frac{690\times 10^{-9}}{sin(23^{\circ})}$
$W=\frac{690\times 10^{-9}}{0.39073}$
$W=1.7\times 10^{-6}m$
$W=1.7\mu m$
