A pole stands vertically inside a triangular park $$ABC$$. Let the angle of elevation of the top of the pole from each corner of the park be $$\frac{\pi}{3}$$. If the radius of the circumcircle of $$\triangle ABC$$ is 2, then the height of the pole is equal to :

A

$$\frac{2\sqrt{3}}{3}$$

B

$$2\sqrt{3}$$

C

$$\sqrt{3}$$

D

$$\frac{1}{\sqrt{3}}$$