Using Snell's law across parallel layers from initial water medium to the final polarizing interface:
$$n_w \sin i = n_w \sin r_{\text{water layer}}$$
Let $$r_w$$ be the angle of incidence at the final interface between the second water layer ($$n_w$$) and the upper medium ($$n_p$$). By geometry:
$$n_w \sin i = n_w \sin r_w$$
Using Brewster's law for complete polarization at the final $$n_w \to n_p$$ reflection interface:
$$\tan r_w = \frac{n_p}{n_w}$$
$$\tan r_w = \frac{4/\sqrt{3}}{4/3} = \frac{3}{\sqrt{3}} = \sqrt{3} \implies r_w = 60^\circ$$
$$\sin i = \sin r_w \implies i = r_w = 60^\circ$$