To find the value of the angle of the prism given the refractive index of the prism $$\mu = \sqrt{3}$$ and the ratio of the angle of minimum deviation ($$\delta_m$$) to the angle of the prism ($$A$$) is 1 (i.e., $$\frac{\delta_m}{A} = 1 \Rightarrow \delta_m = A$$), we can use the prism formula relating these variables.

The formula that relates the angle of deviation $$\delta$$, refractive index $$\mu$$, angle of prism $$A$$, and angle of minimum deviation $$\delta_m$$ is given by:

$$\mu = \frac{\sin \left(\frac{\delta_m + A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$.

Given that $$\delta_m = A$$, our formula becomes:

$$\sqrt{3} = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$

Simplifying this, we have:

$$\sqrt{3} = \frac{\sin(A)}{\sin \left(\frac{A}{2}\right)}$$

Now, recall the trigonometric identity:

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$

By setting $$\theta = \frac{A}{2}$$, we get:

$$\sin(A) = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$$

Substituting back into our equation:

$$\sqrt{3} = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$

$$\sqrt{3} = 2\cos\left(\frac{A}{2}\right)$$

Dividing by 2 and solving for $$\cos\left(\frac{A}{2}\right)$$:

$$\cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2}$$

This corresponds to an angle $$\frac{A}{2}$$ of $$30^\circ$$ since the cosine of 30 degrees is $$\frac{\sqrt{3}}{2}$$. Thus:

$$\frac{A}{2} = 30^\circ$$

So, the angle of the prism $$A$$ is:

$$A = 2 \times 30^\circ = 60^\circ$$.

Therefore, the value of the angle of the prism is $$60^\circ$$.