Using Snell's law we can write
[tex]n_1\sin\theta_1=\text{ n}_2\sin\theta_2;[/tex]Here
[tex]\begin{gathered} n_1=\text{ refractive index of incident medium= 1.46;} \\ n_2=\text{ refractive index of refracting medium = 1.33} \\ \theta_1=\text{ angle of incident= 35}\degree; \\ \theta_2=\text{ angle of refraction} \end{gathered}[/tex]S0 according to problem
[tex]\begin{gathered} 1.46\sin35\degree\text{ = 1.33}\sin\theta_2; \\ sin\theta_2=\text{ }\frac{1.46\times0.57638}{1.33}\begin{cases}sin35={0.57638} \\ \end{cases} \\ sin\theta_2=0.6327; \\ \theta_{2=\text{ }}\sin^{-1}(0.6327); \end{gathered}[/tex]Final answer is
[tex]\sin^{-1}(0.6327)[/tex]