Let me take a shot

The concept is angle is subtended at the center is twice as much as angle subtented by chord. Since angle ABF is 30 deg, angle ACF would be 60 deg and since angle BFG is 30 deg, angle BCG is 60 deg.

Now take a look at the attachment. The area of the shaded portion = Area of circle (pi * 2^2=4pi)- (Area of semicircle AB(2pi) + Area of sector ACF (pi * 2^2 * 60/360 = 2pi/3)+ Area of sector BCG (pi * 2^2 * 60/360 = 2pi/3) + Area of equilateral triangle CFG ( 3^(1/2)/4 * 2^2)

Now Area of shaded portion = 4pi - (2pi+ 2* 2pi/3+ 3^1/2)

= 2/3 pi- 3^(1/2) Ans A