Answer
$m=11$
Work Step by Step
There are no refraction angles greater than $90^{\circ}$, so we can solve for $m_{\max }$
$
m_{\max }=\frac{d \sin 90^{\circ}}{\lambda_2}=\frac{d}{\lambda_2}=\frac{0.0056 \mathrm{~mm}}{5.0 \times 10^{-4} \mathrm{~mm}} \approx 11
$
