ABCD is a square and the arcs centered at the vertices of the square and the radii, are all equal to the side-lengths of the square, (=a). I feel like the arc lengths should all measure 90 degrees as they are the arcs subtended by the angle at each vertex. I'm not sure that helps any, but I've never seen anything like this. My prof gave us a hint saying that we should first try showing the arcs are 30 degrees (not sure where that came from) and that the area of a sector of a disc corresponding to an angle s(in radians) is given by (sR^2)/2
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.
$\square$