Answer
$40^{\circ}$
Work Step by Step
We know that
$n_{air}sin\theta_i=n_wsin\theta_2$
This can be rearranged as:
$\theta_i=sni^{-1}(\frac{n_w}{n_{air}}sin\theta_2)$
We plug in the known values to obtain:
$\theta=sin^{-1}(\frac{1.33}{1.00}sin^135{\circ})=50^{\circ}$
Now we can find the required angle as
$\theta=90^{\circ}-\theta_i=90^{\circ}-50^{\circ}=40^{\circ}$
