Note that $ABCD$ is a parallelogram, then $AD//BC$. Hence, $AF// BG$.
Consider $\Delta DEF$ and $\Delta CEB$. Since $\Delta DEF \sim \Delta CEB$, then we have
$\begin{array}{rcl}
\dfrac{\text{the area of $\Delta DEF$}}{\text{the area of $\Delta CEB$}} & = & \left( \dfrac{DE}{CE} \right)^2 \\
\dfrac{8}{\text{the area of $\Delta CEB$}} & = & \left( \dfrac{2}{3} \right)^2 \\
\dfrac{8}{\text{the area of $\Delta CEB$}} & = & \dfrac{4}{9} \\
\text{the area of $\Delta CEB$} & = & 18 \text{ cm}^2
\end{array}$
Note that $ABCD$ is a parallelogram, then $AB//DC$.
Consider $\Delta ADE$ and $\Delta BCE$. Since $\Delta ADE$ and $\Delta BCE$ have the same height with bases $DE$ and $CE$, then we have
$\begin{array}{rcl}
\dfrac{\text{the area of $\Delta ADE$}}{\text{the area of $\Delta BCE$}} & = & \dfrac{DE}{CE} \\
\dfrac{\text{the area of $\Delta ADE$}}{18} & = & \dfrac{2}{3} \\
\text{the area of $\Delta ADE$} & = & 12 \text{ cm}^2
\end{array}$
Consider $\Delta DEA$ and $\Delta CEG$. Since $\Delta DEA \sim \Delta CEG$, then we have
$\begin{array}{rcl}
\dfrac{\text{the area of $\Delta DEA$}}{\text{the area of $\Delta CEG$}} & = & \left( \dfrac{DE}{CE} \right)^2 \\
\dfrac{12}{\text{the area of $\Delta CEG$}} & = & \left( \dfrac{2}{3} \right)^2 \\
\dfrac{12}{\text{the area of $\Delta CEG$}} & = & \dfrac{4}{9} \\
\text{the area of $\Delta CEG$} & = & 27 \text{ cm}^2
\end{array}$
