Answer
$5.5cm$
Work Step by Step
We can find the required diameter as follows:
$\theta_{min}=2.5 arc \space seconds=(\frac{2.5}{3600})(\frac{2\pi rad}{360^{\circ}})=0.121\times 10^{-4}rad$
Now $D=1.22(\frac{\lambda}{\theta_{min}})$
We plug in the known values to obtain:
$D=1.22(\frac{550\times 10^{-9}m}{0.121\times 10^{-4}m})$
$D=0.055m$
$D=5.5cm$