A parallel sides slab $ABCD$ of refractive index $2$ is sandwiched between two slabs of refractive indices $\sqrt{2}$ and $\sqrt{3}$. The minimum value of angle $\theta$ such that the ray $PQ$ suffers total internal reflection at both the surfaces $AB$ and $CD$ is

For total internal reflection at surface $AB$, angle $\theta$ must be greater than or equal to the critical angle $i_1$
given by

 $\sin i_1=\frac{{\mu }_2}{{\mu }_1}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

which gives $i_1={45}^o$
For total internal reflection at surface $CD$, angle $\theta$ must be greater than or equal to the critical angle $i_2$ given by

$\sin i_2=\frac{{\mu }_3}{{\mu }_1}=\frac{\sqrt{3}}{2}$

which gives $i_2={60}^o$.

Hence, for total internal reflection at both the surfaces $AB$ and $CD$, the minimum value of $\theta ={60}^o$.