$$Area\ of\ equilateral\ triangle\ =\ \frac{\sqrt{3}}{4}a^2$$
Where a = the side of the triangle
$$\sqrt{3}=\ \frac{\sqrt{3}}{4}a^2$$
$$\sqrt{3}=\ \frac{\sqrt{3}}{4}\ \cdot\ a^2$$
$$\frac{\left(\sqrt{3}\ \cdot\ 4\right)}{\sqrt{3}}=\ \frac{\left(\sqrt{3}\ \cdot\ a^2\right)}{\sqrt{3}}$$
$$4\ =\ a^2$$
a = 2

Radius of the inscribed circle in the equilateral triangle,
$$=\ \frac{2a}{p}$$
Where = Area of triangle and p = perimeter of triangle.
$$r\ =\ \frac{\left(2\ \cdot\ \sqrt{3}\right)}{2\ +\ 2\ 2}\ =\ \frac{\left(2\sqrt{3}\right)}{6}$$
$$rationalizing\ the\ surds\ r=\ \frac{\left(2\sqrt{3}\ \cdot\ 2\sqrt{3}\right)}{6\ \cdot2\ \sqrt{3}}\ $$
$$\ r=\ \frac{\left(2\ \cdot\ 2\ \cdot\ \sqrt{3}\right)}{6\ \cdot2\ \cdot\sqrt{3}}\ =\ \frac{\left(4\sqrt{9}\right)}{12\sqrt{3}}=\frac{\left(4\ \cdot3\right)}{12\sqrt{3}}=\frac{12}{12\sqrt{3\sqrt{ }}}$$
$$r\ =\ \frac{1}{\sqrt{3}}$$
$$Area\ of\ inscribed\ circle\ =\ \pi r^2\ =\ \pi\ \cdot\ \left(\frac{1}{\sqrt{3}}\right)^2$$
$$a\ =\ \frac{\pi}{3}$$
$$Area\ of\ shaded\ region\ =\ Area\ of\ triangle\ -\ Area\ of\ inscribed\ circle.$$
$$=\ \sqrt{3}-\frac{\pi}{3}\ =\ Option\ D$$

Option D is CORRECT.