smc

Divide the equilateral hexagon $ABCDEF$ into isosceles triangles $ABF$, $CBD$, and $EDF$ and triangle $BDF$. The three isosceles triangles are congruent by SAS congruence. By CPCTC, $BF=BD=DF$, so triangle $BDF$ is equilateral. Let the side length of the hexagon be $s$. The area of each isosceles triangle is $$\frac{1}{2}ab\sin \angle C=\frac{1}{2}\cdot s\cdot s\cdot \sin 30^{\circ }=\frac{1}{4}{s}^2.$$ By the Law of Cosines on triangle $ABF$, $$B{F}^2=s^2+s^2-2s^2\cos 30^{\circ }=2s^2-\sqrt{3}{s}^2.$$ Hence, the area of the equilateral triangle $BDF$ is $$\frac{\sqrt{3}}{4}B{F}^2=\frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}{s}^2\right)=\frac{\sqrt{3}}{2}{s}^2-\frac{3}{4}{s}^2.$$ The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or $$3\left(\frac{1}{4}{s}^2\right)+\left(\frac{\sqrt{3}}{2}{s}^2-\frac{3}{4}{s}^2\right)=\frac{\sqrt{3}}{2}{s}^2=6\sqrt{3}.$$ Hence, $s=2\sqrt{3}$, and the perimeter of the hexagon is $6s=\boxed{12\sqrt{3}}$.