Answer
$n_3 = 1.0$
Work Step by Step
We can use Equation (33-49) to find the Brewster angle $\theta_B$ at the interface between air $n_1$ and material 2 $n_2$:
$\theta_B = tan^{-1}~\frac{n_2}{n_1}$
$\theta_B = tan^{-1}~\frac{1.5}{1.00}$
$\theta_B = 56.3^{\circ}$
Note that $\theta_1 = \theta_B$
We can find use Snell's law to find $\theta_2$, the angle of refraction as the ray enters material 2:
$n_2~sin~\theta_2 = n_1~sin~\theta_1$
$sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$
$\theta_2 = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_2})$
$\theta_2 = sin^{-1}~(\frac{1.00~sin~56.3^{\circ}}{1.5})$
$\theta_2 = sin^{-1}~(0.55464)$
$\theta_2 = 33.7^{\circ}$
Note that $\theta_2$ is the incident angle as the ray enters material 3. It is given that this angle is the Brewster angle.
We can find $n_3$:
$\theta_2 = tan^{-1}~\frac{n_3}{n_2}$
$tan~\theta_2 = \frac{n_3}{n_2}$
$n_3 = n_2~tan~\theta_2$
$n_3 = 1.5~tan~33.7^{\circ}$
$n_3 = 1.0$
