Answer
Ray $a$ would emerge at an angle of $0^{\circ}$
Work Step by Step
We can use Snell's law to find $\theta_2$:
$n_1~sin~\theta_1 = n_2~sin~\theta_2$
$sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$
$sin~\theta_2 = \frac{n_1~sin~0^{\circ}}{n_2}$
$sin~\theta_2 = 0$
$\theta_2 = 0^{\circ}$
Similarly, as the ray passes across each boundary, the refractive angle will continue to be $0^{\circ}$ in each material.
Ray $a$ would emerge at an angle of $0^{\circ}$.
