smc

Let the given triangle be $△ABE$ with $AB=18$. Also, let $\overline{CD}\parallel \overline{AB}$ with $C$ on $\overline{BE}$ and $D$ on $\overline{AE}$. We know from the problem that $[ABCD]=\frac{1}{3}[△ABE]$, so $[△DCE]=\frac{2}{3}[△ABE]$. Because $\overline{DC}\parallel \overline{AB}$, $△DCE\sim △ABE$ by AA similarity. Because the ratio of these two triangles' areas is $\frac{2}{3}$, the ratio of their sides must be $\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$. Thus, $CD=\frac{\sqrt{6}}{3}AB=\frac{\sqrt{6}}{3}\ast 18=\boxed{6\sqrt{6}}$, which is answer choice $\boxed{6\sqrt{6}}$.