Correct Answer: $\frac{8\pi }{3}-2\sqrt{3}$

Solution :
Given that the area of the intersecting region of two identical circles each of radius $2\;\mathrm{cm}$.
In triangle $\mathrm{AOO', OA = O'A = OO'} = 2\;\mathrm{cm}$
It is an equilateral triangle.
$\angle \mathrm{AOO'} = 60^\circ$
$\angle \mathrm{AOB} = 120^\circ$
In $\triangle \mathrm{ACO}$,
$\mathrm{AC^2 = OA^2 - OC^2} = 4 - 1 = 3$
$\mathrm{AC} = \sqrt{3}$
$\mathrm{AB} = 2\sqrt{3}$
Area of $\mathrm{AO'BA} =$ Area of sector $\mathrm{AO'BO} $$-$ Area of triangle $\mathrm{AOB}$
$=2\times\frac{60^\circ\pi(2)^2}{360^\circ} - \frac{1}{2} \times 2\sqrt{3} \times 1$
$=\frac{\pi (2)^2}{3} - \frac{1}{2} \times 2\sqrt{3} \times 1$
$=\frac{4\pi}{3} - \sqrt{3}$
Area of intersecting region $= 2$ (Area of $\mathrm{AO'BA}$)
$=2 \left(\frac{4\pi}{3} - \sqrt{3}\right) =\left(\frac{8\pi}{3} - 2\sqrt{3}\right)\;\mathrm{cm^2}$
Hence, the correct answer is $\frac{8\pi }{3}-2\sqrt{3}$.