Answer
$10.6^{\circ}$
Work Step by Step
We can find the required angular size of the moon as follows:
$\theta^{\circ}=(\frac{f_{obj}}{f_e})\theta $
We plug in the known values to obtain:
$\theta^{\prime}=(\frac{53cm}{2.5cm})(0.50^{\circ})$
$\theta^{\prime}=10.6^{\circ}$
