As per the question, it is given that
Area of square inscribed the circle $=64c{{m}^{2}}$
$sid{{e}^{2}}=64$
Side $=8cm$
Thus, $AB=BC=CD=DA=8cm$
Using Pythagoras theorem in triangle ABC,
$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$
$A{{C}^{2}}={{8}^{2}}+{{8}^{2}}$
$A{{C}^{2}}=64+64=128$
$AC=\sqrt{128}=8\sqrt{2}cm$
Then, as $\angle B={{90}^{\circ }}$ and AC being the diameter of the circle
The radius is $AC/2=8\sqrt{2}/2=4\sqrt{2}cm$
therefore, the area of the circle $=\pi {{r}^{2}}=3.14{{\left( 4\sqrt{2} \right)}^{2}}$
$=100.48c{{m}^{2}}$