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