Answer
Ray $b$ emerges at an angle of $~~13^{\circ}$
Work Step by Step
Let $\theta_f$ be the angle at which ray $b$ emerges.
According to Snell's law:
$n_{air}~sin~\theta_b = n_1~sin~\theta_1 = n_2~sin~\theta_2 =... = n_{glass}~sin~\theta_f$
$n_{air}~sin~\theta_b = n_{glass}~sin~\theta_f$
$sin~\theta_f = \frac{n_{air}~sin~\theta_b}{n_{glass}}$
$\theta_f = sin^{-1}~(\frac{n_{air}~sin~\theta_b}{n_{glass}})$
$\theta_f = sin^{-1}~(\frac{1.0~sin~20^{\circ}}{1.5})$
$\theta_f = sin^{-1}~(0.228)$
$\theta_f = 13^{\circ}$
Ray $b$ emerges at an angle of $~~13^{\circ}$.
