Answer
$\theta_B = 51.1^{\circ}$
Work Step by Step
We can use Equation (33-45) to find the critical angle $\theta_c$:
$\theta_c = sin^{-1}~\frac{n_3}{n_2}$
$\theta_c = sin^{-1}~\frac{1.30}{1.80}$
$\theta_c = 46.24^{\circ}$
Note that the refracted angle in material 2 is $90^{\circ}-\theta_c$ which is $43.76^{\circ}$
We can use Snell's law to find $\theta_B$:
$n_1~sin~\theta_B = n_2~sin~\theta_2$
$sin~\theta_B = \frac{n_2~sin~\theta_2}{n_1}$
$\theta_B = sin^{-1}~(\frac{n_2~sin~\theta_2}{n_1})$
$\theta_B = sin^{-1}~(\frac{1.80~sin~43.76^{\circ}}{1.60})$
$\theta_B = sin^{-1}~(0.7781)$
$\theta_B = 51.1^{\circ}$
