imc

We draw a diagram to make our work easier: Assume that $AB$ is of length $1$. Therefore, the area of $ABCDEF$ is $\frac{3\sqrt{3}}{2}$. To find the area of $WCXYFZ$, we draw $\overline{CF}$, and find the area of the trapezoids $WCFZ$ and $CXYF$. From this, we know that $CF=2$. We also know that the combined heights of the trapezoids is $\frac{\sqrt{3}}{3}$, since $\overline{ZW}$ and $\overline{YX}$ are equally spaced, and the height of each of the trapezoids is $\frac{\sqrt{3}}{6}$. From this, we know $\overline{ZW}$ and $\overline{YX}$ are each $\frac{1}{3}$ of the way from $\overline{CF}$ to $\overline{DE}$ and $\overline{AB}$, respectively. We know that these are both equal to $\frac{5}{3}$. We find the area of each of the trapezoids, which both happen to be $\frac{11}{6}\cdot \frac{\sqrt{3}}{6}=\frac{11\sqrt{3}}{36}$, and the combined area is $\frac{11\sqrt{3}}{18}$. We find that $\frac{\frac{11\sqrt{3}}{18}}{\frac{3\sqrt{3}}{2}}$ is equal to $\frac{22}{54}=\boxed{\frac{11}{27}}$.