Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1007: 57

$5.5cm$

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: $D=1.22(\frac{550\times 10^{-9}m}{0.121\times 10^{-4}m})$ $D=0.055m$ $D=5.5cm$