1. Condition at minimum deviation

For a prism at minimum deviation,

• angle of incidence equals angle of emergence: $i=e$
• angle of refraction at both faces are equal: $r_1=r_2=r$

Hence, for prism angle $A$, $A=r_1+r_2=2r$ so, $r=\frac{A}{2}$

2. Given relation

It is given that the angle of incidence is double the angle of refraction inside the prism: $i=2r$

3. Apply Snell's law at first face

For air to prism, $\mu=\frac{\sin i}{\sin r}$ Given $\mu=\sqrt{3}$ and $i=2r$: $\sqrt{3}=\frac{\sin 2r}{\sin r}$

Using $\sin 2r=2\sin r\cos r$, $\sqrt{3}=\frac{2\sin r\cos r}{\sin r}=2\cos r$

Therefore, $\cos r=\frac{\sqrt{3}}{2}$

So, $r=30^\circ$

4. Find prism angle

Since, $A=2r=2\times 30^\circ=60^\circ$

5. Final answer

The angle of the prism is $\boxed{60^\circ}$