Answer
(a) $40.6^{\circ}$
(b) The answer to part (a) doesn't depend on the thickness of the oil film.
Work Step by Step
(a) We know that
$\theta_w=sin^{-1}(\frac{n_a}{n_w}sin\theta_a)$
We plug in the known values to obtain:
$\theta_w=sin^{-1}(\frac{1.00}{1.33}sin60.0^{\circ})=40.6^{\circ}$
(b) We know that the answer to part (a) doesn't depend on the thickness of the oil film. It only depends on the original angle of incidence and the refractive index of water and air.