smc

Let the center of the hexagon be $O$. $△AOB$, $△BOC$, $△COD$, $△DOE$, $△EOF$, and $△FOA$ are all equilateral triangles with side length $2$. Thus, $CO=2$, and $GO=\sqrt{A{O}^2-A{G}^2}=\sqrt{3}$. By symmetry, $\angle COG=90^{\circ }$. Thus, by the Pythagorean theorem, $CG=\sqrt{2^2+{\sqrt{3}}^2}=\sqrt{7}$. Because $CO=OF$ and $GO=OH$, $CG=HC=FH=GF=\sqrt{7}$. Thus, the solution to our problem is $\sqrt{7}+\sqrt{7}+\sqrt{7}+\sqrt{7}=\boxed{4\sqrt{7}}$.