Since the inscribed triangle is equilateral, therefore the angles at all the points = 60°

Using the formula for inscribed circle, 

2R = \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

where R = radius of the circle; a, b and c are the sides of the triangle.

⇒ 2R = \(\frac{\sqrt{3}}{\sin 60}\)

 2R = \(\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}\)

2R = 2

R = 1cm