Answer: E

Solution:

Let the common side length be $s.$ The three acute vertices are the tips of isosceles triangles with two sides $s$ and apex $30^\circ;$ each has area $\tfrac12 s^2 \sin 30^\circ = \tfrac{s^2}{4}.$

The three reflex vertices form an inner equilateral triangle with side $2s\sin 15^\circ,$ whose area is $\sqrt3\,s^2\sin^2 15^\circ.$ Using $\sin^2 15^\circ = \tfrac{2 - \sqrt3}{4},$ the total area is $$\frac{3s^2}{4} + \sqrt3\,s^2\cdot\frac{2 - \sqrt3}{4} = \frac{s^2\sqrt3}{2}.$$

Setting $\tfrac{s^2\sqrt3}{2} = 6\sqrt3$ gives $s^2 = 12,$ so $s = 2\sqrt3$ and the perimeter is $6s = 12\sqrt3.$

Thus, the correct answer is E.