We may and do assume that $a=1$. (Multiply the result for $1$ with $a^2$.)
From the given data, let us make a picture containing only the given square $ABCD$, and the two equilateral triangles $\Delta ABH$, and $BCG$.
From
$
\widehat{CBG}=
\widehat{HBA}=60^\circ$, we obtain three angles of $30^\circ$
from $\widehat{CBA}$, separated by the rays $BG$, and $BH$.
So the area of the disk sector of the disk centered in $B$, bounded by $BH$ and $BG$ is
$$
\pi\cdot\frac{30^\circ}{360^\circ}=\frac\pi{12}\ .
$$
The area of the triangle $\Delta BGH$ is
$$
[BGH]=\frac 12\cdot BG\cdot BH\cdot\sin 30^\circ=\frac 14\ .
$$
Note that the difference is the "small" value $\displaystyle\frac 1{12}(\pi-3)$, the area between the chord $GH$ and the arc $\overset\frown{GH}$. Such an area comes four times in the final area, so this contribution is
$$
4\cdot
\frac1{12}(\pi-3)
=
\frac13(\pi-3)\ .
$$
Else there remains the area of a square with side $GH$.
We can compute $GH$ from the isosceles triangle $\Delta BGH$, by drawing its height and angle bisector in $B$. We obtain $GH=2\sin 15^\circ$. The area of the square $EFGH$ is then
$$
\begin{aligned}{}
[EFGH]
&=GH^2
=(2\sin15^\circ)^2
=4\sin^2 15^\circ
=4 - 4\cos^2 15^\circ
\\
&= 4-2(1+\cos30^\circ) = 4-2-2\cdot \frac {\sqrt3}2\\
&=2-\sqrt 3
\ .
\end{aligned}
$$
It remains to add the two areas.
$\square$
