From the figure below

$\angle DAB = 30^\circ$, $\angle EBA = 30^\circ$, $\angle ABD = 60^\circ$

$r^2 + 10^2 = 20^2$

$r^2 = 300$

$r = 17.32 \, \text{ cm}$

$BF = AD = r = 17.32 \, \text{ cm}$

$EF^2 + 10^2 = r^2$

$EF^2 + 100 = 300$

$EF = 14.14 \, \text{ cm}$

$BE = BF - EF = 17.32 - 14.14$

$BE = 3.18 \, \text{ cm}$

Sine law for triangle ABE

$\dfrac{r}{\sin \angle EBA} = \dfrac{BE}{\sin \angle EAB}$

$\dfrac{17.32}{\sin 30^\circ} = \dfrac{3.18}{\sin \angle EAB}$

$\sin \angle EAB = \dfrac{3.18 \sin 30^\circ}{17.32}$

$\angle EAB = 5.27^\circ$

$\angle DAE = \angle DAB - \angle EAB = 30^\circ - 5.27^\circ$

$\angle DAE = 24.73^\circ$

$A_{BED} = A_{ABD} - A_{ABE} - A_{AED}$

$A_{BED} = \frac{1}{2}(10)(20)\sin 60^\circ - \frac{1}{2}(20)(3.18) \sin 30^\circ - \dfrac{\pi(17.32^2)(24.73^\circ)}{360^\circ}$

$A_{BED} = 86.60 - 15.9 - 64.74$

$A_{BED} = 5.96 \, \text{ cm}^2$

Required Area

$A_{required} = A_{ABC} - 6A_{BED}$

$A_{required} = \frac{1}{2}(20^2) \sin 60^\circ - 6(5.96)$

$A_{required} = 137.44 \, \text{ cm}^2$ *answer*