Consider the triangle made by the center of the two circular arcs and their intersection.
This triangle is equilateral, see
.
Its side length is \(1\) so its area is \(\dfrac{\sqrt{3}}{4}\).
The two circular sectors outside the triangle have radius \(1\).
Since the angles of the equilateral triangle are \(60^\circ\), the central angle of the circular sectors is \(30^\circ\).
Thus, each circular sector is the twelfth of a circle and it has area \(\dfrac{\pi}{12}\).
The marked region can be obtain from the square by removing the triangle and the two circular sectors.
Its area is
\[1-\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{6}\,.\]