1. Use the relation for critical angle

For light going from a denser medium $(n_d)$ to a rarer medium $(n_r)$, the critical angle $C$ satisfies:

$\sin C = \frac{n_r}{n_d}$

Given:

$C = 45^\circ$

So,

$\sin 45^\circ = \frac{n_r}{n_d} = \frac{1}{\sqrt{2}}$

Thus,

$\frac{n_r}{n_d} = \frac{1}{\sqrt{2}} \Rightarrow \frac{n_d}{n_r} = \sqrt{2}$

2. Relate refractive index to speed

Since refractive index is inversely proportional to speed of light in the medium:

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

Therefore,

$\frac{n_d}{n_r} = \frac{v_r}{v_d}$

So,

$\sqrt{2} = \frac{v_r}{v_d}$

Given speed in rarer medium:

$v_r = 3 \times 10^8\ \text{m/s}$

Hence,

$v_d = \frac{v_r}{\sqrt{2}} = \frac{3 \times 10^8}{\sqrt{2}}$

$v_d = 2.12 \times 10^8\ \text{m/s}$

3. Match with the options

This corresponds to:

Option A: $2.12 \times 10^8\ \text{m/s}$

4. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They match.