Answer: E

Solution:

Each interior angle is $120^\circ,$ so extending sides $FA$ and $BC,$ $BC$ and $DE,$ and $DE$ and $FA$ cuts off three equilateral corner triangles and forms a large equilateral triangle.

Let $EF=e$ and $FA=f.$ Resolving the six sides in directions separated by $60^\circ$ gives $f-e=4$ and $e+f=6,$ so $e=1$ and $f=5.$ The corner triangles built on $AB, CD,$ and $EF$ are therefore equilateral. The large triangle has side $1+4+2=7,$ while the removed triangles have sides $1,2,$ and $1.$ The area is $$\begin{aligned} &\frac{\sqrt3}{4}\left(7^2 - 1^2 - 2^2 - 1^2\right) \\ &\quad = \frac{43\sqrt3}{4}. \end{aligned}$$

Thus, the correct answer is E.