Answer
$2.1m$
Work Step by Step
We can find the required diameter as follows:
$D=\frac{1.22\lambda}{tan^{-1}(\frac{y}{L})}$
We plug in the known values to obtain:
$D=\frac{1.22(550\times 10^{-9}m)}{tan^{-1}(\frac{0.05m}{160\times 10^3m})}$
$D=214\times 10^{-3}m$
$D=2.1m$
