Solution:

The answer is (E). The radius of the circle circumscribed about the square is $\frac{\sqrt{2}}{2}$, being half of the diagonal of the square itself. $LMNO$ is also a square, since the figure is symmetric under rotations of $90$ degrees. Now, the vertices of the unit square divide the circle into four arcs. The midpoints of these have the same distance from the sides of the square as the points $L, M, N, O$, being in fact symmetric to them with respect to the respective sides. This distance turns out to be the difference between the radius of the circle and half the side of the square, hence $\frac{\sqrt{2}-1}{2}$. $LN$ equals the difference between the side of the unit square and twice the above distance, that is $2-\sqrt{2}$. Finally, $LN$ is the diagonal of $LMNO$, whose side is therefore $\sqrt{2}-1$, and whose area is $(\sqrt{2}-1)^2=3-2\sqrt{2}$.