jmc

Clearly, since $\overline{FE}$ is a side of a square with area $32$, $\overline{FE}=\sqrt{32}=4\sqrt{2}$. Now, since $\overline{FE}=\overline{BC}$, we have $\overline{BC}=4\sqrt{2}$. Now, $\overline{JB}$ is a side of a square with area $18$, so $\overline{JB}=\sqrt{18}=3\sqrt{2}$. Since $\Delta JBK$ is equilateral, $\overline{BK}=3\sqrt{2}$. Lastly, $\Delta KBC$ is a right triangle. We see that $\angle JBA+\angle ABC+\angle CBK+\angle KBJ=360^{\circ }\to 90^{\circ }+120^{\circ }+\angle CBK+60^{\circ }=360^{\circ }\to \angle CBK=90^{\circ }$, so $\Delta KBC$ is a right triangle with legs $3\sqrt{2}$ and $4\sqrt{2}$. Now, its area is $\frac{3\sqrt{2}\cdot 4\sqrt{2}}{2}=\frac{24}{2}=\boxed{12}$.