Hint: In the case of minimum deviation, the ray of light is parallel to the base of the prism.

Step 1: Use the expression of the refractive index of a prism.

The relationship between refractive index, prism angle A and angle of minimum deviation is given by,
                   $\mathrm{\mu }=\frac{\sin \left[{\displaystyle \frac{(\mathrm{A}+{\mathrm{D}}_{\mathrm{m}})}{2}}\right]}{\sin \left({\displaystyle \frac{\mathrm{A}}{2}}\right)}$
Here,
$\therefore$Given,                              ${\mathrm{D}}_{\mathrm{m}}=\mathrm{A}$
Substituting the value, we have,
$\therefore$                                             $\mathrm{\mu }=\frac{\sin \mathrm{A}}{\sin {\displaystyle \frac{\mathrm{A}}{2}}}$
On solving, we have                     $\mathrm{\mu }=\frac{2\sin {\displaystyle \frac{\mathrm{A}}{2}}\cos {\displaystyle \frac{\mathrm{A}}{2}}}{\sin {\displaystyle \frac{\mathrm{A}}{2}}}=2 \cos \frac{\mathrm{A}}{2}$

Step 2: Find the value of the angle of the prism.
For the given value of the refractive index,
we have.
$\therefore$                                            $\cos \frac{\mathrm{A}}{2}=\frac{\sqrt{3}}{2}$
or                                                  $\frac{\mathrm{A}}{2}=30°$
$\therefore$                                                  A=60$°$
This is the required value of the prism angle.