Correct Answer: $72\sqrt{3}$

Solution : Given that ABCDEF is a regular hexagon with a side length of 6 cm, and APF, QAB, DCR, and DES are equilateral triangles.
The area of an equilateral triangle = $\frac{\sqrt{3}}{4} s^2 $ where $s$ is the side length of the triangle.
On substituting $s$ = 6 cm.
The area of an equilateral triangle APF = $\frac{\sqrt{3}}{4} (6)^2 = 9\sqrt{3} $ cm ²
Area of quadrilateral ABOF = 2 × Area of an equilateral triangle APF
⇒ Area of quadrilateral ABOF = $18\sqrt{3} $ cm ²
Similarly, of the area of quadrilateral DCOE = 2 × Area of an equilateral triangle EDS
⇒ Area of quadrilateral DCOE = $18\sqrt{3} $ cm ²
The area of the shaded region = Area of quadrilateral ABOF + Area of quadrilateral DCOE + 4 × Area of an equilateral triangle APF
⇒ The area of the shaded region = $18\sqrt{3} +18\sqrt{3} +36\sqrt{3} =72\sqrt{3} $ cm ²
Hence, the correct answer is $72\sqrt{3} $.