Area of shaded region

Considering only one part shaded portion (enlarged figure) and let $A_s$ be the area of one shaded portion

$\large A_{sector}=\dfrac{1}{4} \pi a^2=\dfrac{\pi}{4}a^2$

$\large A_{segment}=a^2-\dfrac{\pi}{4}a^2$

$\large A_s=a^2-2\left(a^2-\dfrac{\pi}{4}a^2 \right)=a^2-2a^2-\dfrac{\pi}{4}a^2=-a^2+\dfrac{\pi}{2}a^2$

Since there are $4$ $A_s$ , we have,

$\large 4A_s=4\left(-a^2+\dfrac{\pi}{2}a^2 \right)$

$\large 4A_s=-4a^2+2\pi a^2$

$\large 4A_s=a^2(-4+2\pi)$

$\large 4A_s=$ $\color{plum}\boxed{\large a^2(2\pi - 4)}$

The area is composed of 8 circular segments, each of them a quarter circle with radius $a$ , area $\frac14\pi a^2$ , minus a right triangle with both legs equal $a$ , area $\frac12 a^2$ .

So the total area is $8\times(\frac{\pi}4 a^2-\frac12 a^2)=\boxed{(2\pi-4)a^2}$