Answer
(B)$~1.052\times 10^{-4}rad$
Work Step by Step
We know that
$\theta_{min}=\frac{1.22\lambda}{Dn}$
We plug in the known values to obtain:
$\theta_{min}=\frac{1.22(645\times 10^{-9}m)}{(5.50\times 10^{-3}m)(1.36)}$
$\theta_{min}=1.052\times 10^{-4}rad$