Solution:

Given that the three circles are drawn such that each of them touches the other two.

Now, by joining the centers of the three circles,

We get,  AB = BC = CA = \[2\] (radius) = \[7\] cm

Therefore, we can say that triangle ABC is an equilateral triangle with each side \[7\] cm.

∴ Area of the triangle of side a  = \[(\sqrt{3}/4)\times {{a}^{2}}\]

= \[(\sqrt{3}/4)\times {{7}^{2}}\]

= \[(49/4)\sqrt{3}\] \[c{{m}^{2}}\]

= \[21.2176\] \[c{{m}^{2}}\]

Now, we know that Central angle of each sector = \[\phi ={{60}^{\circ }}(60\pi /180)\]

= \[\pi /3\] radians

Thus, area of each sector = \[(1/2){{r}^{2}}\theta \]

= \[(1/2)\times {{(3.5)}^{2}}\times (\pi /3)\]

= \[12.25\times (22/(7\times 6))\]

= \[6.4167\]\[c{{m}^{2}}\]

Therefore, Total area of three sectors = \[3\times 6.4167\] = \[19.25\] \[c{{m}^{2}}\]

so, the Area enclosed between three circles = Area of triangle ABC – Area of the three sectors

= \[21.2176-19.25\]

= \[1.9676\] \[c{{m}^{2}}\]

Therefore, the required area enclosed between these circles is \[1.9676\] \[c{{m}^{2}}\]