Answer
$25\;\rm \mu F$
Work Step by Step
First, we need to re-sketch this circuit and reduce it as we see in the figures below.
The two capacitors, in the first circuit below, $C_1$ and $C_2$ are in series so their equivalent is given by
$$(C_{eq})_{12 }=\left[ \dfrac{1}{C_1}+\dfrac{1}{C_2} \right]^{-1}$$
Plug the known;
$$(C_{eq})_{12}=\left[ \dfrac{1}{20}+\dfrac{1}{30} \right]^{-1}$$
$$(C_{eq})_{12 }=\color{black}{\bf 12}\;\rm \mu F$$
The two capacitors, in the second circuit below, $(C_{eq})_{23}$ and $C_3$ are in parallel so their equivalent is given by
$$(C_{eq})_{123}=(C_{eq})_{23}+C_3=12+13$$
$$(C_{eq})_{123 }=\color{red}{\bf 25}\;\rm \mu F$$
