From the ray diagram, at point P, using Snell's law: $\frac{\sin i}{\sin r} = \frac{\mu_{air}}{\mu_{prism}}$. Here $i = 30^\circ$ and $\mu_{prism} = \sqrt{3}$. So $\frac{\sin 30^\circ}{\sin e} = \frac{1}{\sqrt{3}}$. $\sin e = \sqrt{3} \cdot \sin 30^\circ = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$. Therefore, $e = 60^\circ$.