Radius of four circles

$r = 30 ~ \text{cm}$

From triangle ECO

$\cos \angle EOC = \dfrac{CE}{OC}$

$\cos 30^\circ = \dfrac{30}{OC}$

$OC = 20\sqrt{3} ~ \text{cm}$

Cosine law for triangle OCD

$OD^2 = CD^2 + OC^2 - 2(CD)(OC) \cos (\frac{1}{2}\theta)$

$30^2 = 30^2 + (20\sqrt{3})^2 - 2(30)(20\sqrt{3}) \cos (\frac{1}{2}\theta)$

$1200 - 1200\sqrt{3} \cos (\frac{1}{2}\theta) = 0$

$\cos (\frac{1}{2}\theta) = 1 / \sqrt{3}$

$\theta = 109.47^\circ$

Area of the circular segment of central angle θ

$A_{segment} = \frac{1}{2}r^2 (\theta_{rad} - \sin \theta_{deg})$

$A_{segment} = \frac{1}{2}(30^2) \left[ 109.47^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 109.47^\circ \right]$

$A_{segment} = 435.52 ~ \text{cm}^2$

Required area

$A = A_{circle} - 6A_{segment}$

$A = \pi (30^2) - 6(435.52)$

$A = 214.32 ~ \text{cm}^2$ *answer*