$\mathit{d}{\mathit{\mu }}_{\mathit{r}}=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$
(where $\mathit{d}{\mathit{\mu }}_{\mathit{r}}$, is the refractive index of rarer medium w.r.t. denser medium)
$\therefore {\mathit{\mu }}_{\mathit{d}}=\sqrt{3}$
$\mathit{v}=\frac{\mathit{c}}{\mathit{\mu }}=\frac{3\times {10}^8}{\sqrt{3}}=\sqrt{3}\times {10}^8\mathrm{m}∕\mathrm{}\mathit{s}$
[Note : Also accept if a student solves it as follows]
$\mathit{\mu }=\tan {\mathit{i}}_{\mathit{p}}$
$\mathit{\mu }=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$

$\therefore \mathit{v}=\frac{3\times {10}^8}{\frac{1}{\sqrt{3}}}=3\sqrt{3}\times {10}^8\mathrm{m}∕\mathrm{}\mathit{s}$