The centers of the three circles, each with radius \(r=2\) cm, form an equilateral triangle because the circles touch each other. The side length of this equilateral triangle is equal to the sum of the radii of two touching circles, which is \(2r=2(2)=4\) cm.
 The area of an equilateral triangle with side length \(a\) is given by the formula \(\frac{\sqrt{3}}{4}a^{2}\). Substituting \(a=4\) cm, the area of the equilateral triangle formed by the centers is \(\frac{\sqrt{3}}{4}(4^{2})=\frac{\sqrt{3}}{4}(16)=4\sqrt{3}\) cm$^2$
. Within this equilateral triangle, there are three circular sectors, one from each circle. Since the angle of an equilateral triangle is \(60^{\circ }\), each sector has a central angle of \(60^{\circ }\). The area of a sector with radius \(r\) and central angle \(\theta \) (in degrees) is given by \(\frac{\theta }{360}\pi r^{2}\). For each sector, the area is \(\frac{60}{360}\pi (2^{2})=\frac{1}{6}\pi (4)=\frac{2}{3}\pi \) cm$^2$. Since there are three such sectors, their total area is \(3\times \frac{2}{3}\pi =2\pi \) cm$^2$.
The area of the region enclosed by the three circles is the area of the equilateral triangle minus the total area of the three circular sectors. Area = Area of equilateral triangle - Area of three sectors Area = \(4\sqrt{3}-2\pi \) cm$^2$.
Option B is given as \(\pi /2(4+2\sqrt{3})^{2}\). Let's expand this expression: \(\pi /2(16+16\sqrt{3}+12)=\pi /2(28+16\sqrt{3})=14\pi +8\pi \sqrt{3}\). This does not directly match the calculated area of \(4\sqrt{3}-2\pi \)