Explanation

(d) Here, $v_{A} = {1.8 \times 10}^{8}ms^{- 1}$
$v_{B} = 2.4 \times 10^{8}ms^{- 1}$
Light travels slower in denser medium. Hence medium $A$ is a denser medium and medium $B$ is a rarer medium. Here, light travels from medium $A$ to medium $B$.
Let $C$ be the critical angle between them
$\therefore\sin{C =}\ ^{A}\mu_{B} = \frac{1}{\ ^{B}\mu_{A}\ }$
Refractive index of medium $B\ w.$r.t medium $A$ is
$\ ^{A}\mu_{B} = \frac{Velocity\ of\ light\ in\ medium\ A}{Velocity\ of\ light\ in\ medium\ B} = \frac{v_{A}}{v_{B}}$
$\therefore\sin C = \frac{v_{A}}{v_{B}} = \frac{1.8 \times 10^{8}}{2.4 \times 10^{8}} = \frac{3}{4}$
$\Rightarrow C = \sin^{- 1}\left( \frac{3}{4} \right)$