As shown in the figure, a ray AB of unpolarized light enters from...

Let the angle of incidence at the first interface (water to glass) be $i$.

By applying Snell's law at the successive parallel interfaces, we can find the angle of incidence at the interface where the reflection occurs (water to medium $n_p$).

For the first interface (water to glass):
$n_w \sin i = n_g \sin r_1$

For the second interface (glass to water):
$n_g \sin r_1 = n_w \sin r_2$

From the above two equations, we get:
$n_w \sin i = n_w \sin r_2 \implies r_2 = i$

Thus, the angle of incidence at the third interface (water to medium $n_p$) is also $i$.

The problem states that the reflected ray is completely polarized with its electric field perpendicular to the plane of incidence (indicated by the dots on ray $CD$). This complete polarization upon reflection occurs when the light is incident at Brewster's angle $\theta_B$.

According to Brewster's law for the interface between water and medium $n_p$:
$\tan \theta_B = \dfrac{n_p}{n_w}$

Since the angle of incidence at this interface is $i$, we have $i = \theta_B$. Substituting the given refractive indices:
$\tan i = \dfrac{4/\sqrt{3}}{4/3}$

$\tan i = \dfrac{3}{\sqrt{3}} = \sqrt{3}$

Therefore, the angle of incidence is:
$i = 60^\circ$

Answer: $60$