Circle Areas

The white region is equal to two two-third circle with radius $6$ plus two equilateral triangles with side length $6$ . Therefore the area is:

$A = 2 \times \frac 23 \times 6^2 \pi + 2 \times \frac {6^2\sin 60^\circ} 2 = \boxed{48\pi + 18\sqrt 3}$

The required area is

$2πr^2-2\left (\dfrac 12\times r^2\times \dfrac {2π}{3}-\dfrac {r^2\sqrt 3}{4}\right )$ ,

where $r=6$ is the radius of each circle.

So, the required area is

$\dfrac {4πr^2}{3}+\dfrac {r^2\sqrt 3}{2}$

$=\boxed {48π+18\sqrt 3}$ .