Solution:

The answer is $\mathbf{(E)}$. The area we are looking for is equal to the area of triangle $ABC$ from which we have subtracted the three smaller triangles that start from the vertices of $ABC$. These triangles are also equilateral (they are homomorphic to $ABC$) and all congruent. We also note that $ABC$ and $DEF$ are congruent. Let us consider, for example, the small triangle with vertex at $A$: its height is equal to the distance from $A$ to the side of triangle $DEF$ parallel to $BC$. By construction, $AO$ is twice this distance and, by the property of the median of a triangle, $AO = OD$ is $\frac{2}{3}$ of the height of triangle $DEF$ (and therefore of $ABC$). Since the small triangle is homomorphic to $ABC$ and has a height $\frac{1}{3}$ that of $ABC$, its area will be $\frac{1}{9}$ that of $ABC$. The area of the three small triangles will therefore be $\frac{1}{3}$ that of $ABC$, and the area we were looking for is $\frac{2}{3}$.