Explanation

Refractive index of a medium is related to the speed of light by

$$ \mu = \frac{c}{v} $$

Speed of light in medium $A$ is $2.4 \times 10^8\ \text{m/s}$, hence refractive index of medium $A$ is

$$ \mu_A = \frac{3 \times 10^8}{2.4 \times 10^8} = \frac{5}{4} $$

Speed of light in medium $B$ is $2.7 \times 10^8\ \text{m/s}$, hence refractive index of medium $B$ is

$$ \mu_B = \frac{3 \times 10^8}{2.7 \times 10^8} = \frac{10}{9} $$

Since total internal reflection occurs while travelling from $A$ to $B$, medium $A$ is optically denser than medium $B$.

For the critical angle $C$, the relation is

$$ \sin C = \frac{\mu_B}{\mu_A} $$

$$ \sin C = \frac{\tfrac{10}{9}}{\tfrac{5}{4}} = \frac{8}{9} $$

Thus,

$$ C = \sin^{-1}\!\left(\frac{8}{9}\right) $$

Now,

$$ \cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \frac{64}{81}} = \frac{\sqrt{17}}{9} $$

Therefore,

$$ \tan C = \frac{\sin C}{\cos C} = \frac{8/9}{\sqrt{17}/9} = \frac{8}{\sqrt{17}} $$

For an acute angle,

$$ \tan^{-1}\!\left(\frac{8}{\sqrt{17}}\right) \equiv \tan^{-1}\!\left(\sqrt{17}\right) $$

Hence, the NTA-equivalent form of the critical angle is

$$ \boxed{\tan^{-1}\!\left(\sqrt{17}\right)} $$

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