Explanation:
Refer to the figure below;

ABCDEF is a regular Hexagon with each angle 120° and AC = 6
Drop a perpendicular (BG) in Isosceles triangle ABC with AB = BC.
BG will bisect the angle ABC and the side AC i.e. AG = CG = 3

Now, ABG and BGC are two congruent 30-60-90 triangles
In triangle BGC, GC = $\sqrt{3}x = 3$

i.e. $x = \frac{3}{\sqrt{3}} = \sqrt{3}$

So, AB = BC = CD = DE = EF = EA = $2x = 2\sqrt{3}$

Things to Remember:
1. A regular Hexagon gets divided into 6 equilateral triangles when all its vertices are joined from the centre (here O)
2. Area of Sector = $\frac{θ}{360}πr^2$
3. Area of an equilateral triangle = $\frac{\sqrt{3}}{4}a^2$

Therefore, DE = EO = OD = $2\sqrt{3}$
Also, radius of the circle O = $2\sqrt{3}$

Area of Shaded region = Area of sector EOD - Area of equilateral triangle EOD
= $\frac{60}{360}π(2\sqrt{3})^2$ - $\frac{\sqrt{3}}{4}(2\sqrt{3})^2$

= $\frac{π}{6}(2\sqrt{3})^2 - \frac{\sqrt{3}}{4}(2\sqrt{3})^2$

= $(\frac{π}{6}-\frac{\sqrt{3}}{4})(2\sqrt{3})^2$

= $\frac{4π-6\sqrt{3}}{24}(12)$

= $\frac{4π-6\sqrt{3}}{2}$

Col. A: $\frac{4π-6\sqrt{3}}{2}$
Col. B: $\frac{3}{2}$

Multiplying by $2$ both sides;

Col. A: $4π-6\sqrt{3}$
Col. B: $3$

Col. A: $2.17$
Col. B: $3$

Hence, option B