Solution:

The answer is (D). The arcs drawn inside the circle have the same radius as the circle itself. As a consequence, by equidecomposition of the circle, the area of the shaded part in figure 1 is equal to that of the shaded part in figure 2, that is, it is equal to the area of the hexagon inscribed in a circle of unit radius. The side of the hexagon is equal to the radius (that is, it equals $1$); the area of the hexagon is $6$ times that of the triangle $OPQ$.

figura 1

figura 2

As a consequence,
$$\mathrm{A}_{\text{esagono}} = 6 \cdot \frac{1}{2} \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{2}.$$