Answer
$\theta=56.3^o$
Work Step by Step
$\theta_2=\arcsin(\frac{n_1\sin(\theta_1)}{n_2})=\arcsin(\frac{(1.0003)\sin(45.0^o)}{1.54})=27.3^o$
$\theta_3$ of ray of light exiting prism can be found because the angles of a triangle formed by the top part of the prism and $90^o-\theta_1$, and $\theta_3$ is $180^o$
$\theta_3=90^o-(180^o-60^o-(90^o-27.3^o))=32.7^o$
$\theta_4=\arcsin(\frac{n_1\sin(\theta_1)}{n_2})=\arcsin(\frac{(1.54)\sin(32.7^o)}{1.0003})=56.3^o$
