smc

Refer to the diagram above. Let the origin be at the center of the square, $A$ be the intersection of the top and right hexagons, $B$ be the intersection of the top and left hexagons, and $M$ and $N$ be the top points in the diagram. By symmetry, $A$ lies on the line $y=x$. The equation of line $AN$ is $y=-x\sqrt{3}+\frac{3}{2}\sqrt{3}-\frac{1}{2}$ (due to it being one of the sides of the top hexagon). Thus, we can solve for the coordinates of $A$ by finding the intersection of the two lines: $$x=-x\sqrt{3}+\frac{3\sqrt{3}-1}{2}$$ $$x(\sqrt{3}+1)=\frac{3\sqrt{3}-1}{2}$$ $$x=\frac{3\sqrt{3}-1}{2}\cdot \frac{1}{\sqrt{3}+1}$$ $$=\frac{3\sqrt{3}-1}{2(\sqrt{3}+1)}\cdot \frac{\sqrt{3}-1}{\sqrt{3}-1}$$ $$=\frac{10-4\sqrt{3}}{4}$$ $$=\frac{5}{2}-\sqrt{3}$$ $$\therefore A=\left(\frac{5}{2}-\sqrt{3},\frac{5}{2}-\sqrt{3}\right).$$ This means that we can find the length $AB$, which is equal to $2(\frac{5}{2}-\sqrt{3})=(5-2\sqrt{3}$. We will next find the area of trapezoid $ABMN$. The lengths of the bases are $1$ and $5-2\sqrt{3}$, and the height is equal to the $y$-coordinate of $M$ minus the $y$-coordinate of $A$. The height of the hexagon is $\sqrt{3}$ and the bottom of the hexagon lies on the line $y=\frac{1}{2}$. Thus, the $y$-coordinate of $M$ is $\sqrt{3}-\frac{1}{2}$, and the height is $2\sqrt{3}-3$. We can now find the area of the trapezoid: $$[ABMN]=(2\sqrt{3}-3)\left(\frac{1+5-2\sqrt{3}}{2}\right)$$ $$=(2\sqrt{3}-3)(3-\sqrt{3})$$ $$=6\sqrt{3}+3\sqrt{3}-9-6$$ $$=9\sqrt{3}-15.$$ The total area of the figure is the area of a square with side length $AB$ plus four times the area of this trapezoid: $$\text{Area}=(5-2\sqrt{3}{)}^2+4(9\sqrt{3}-15)$$ $$=37-20\sqrt{3}+36\sqrt{3}-60$$ $$=16\sqrt{3}-23.$$ Our answer is $16+3-23=\boxed{-4}$.