Answer
$7.5\;\rm \mu F$
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, $C_2$ and $C_3$ are in parallel so their equivalent is given by
$$(C_{eq})_{23}=C_2+C_3=20+10$$
$$(C_{eq})_{23}=\bf 30\;\rm \mu F\tag 2$$
The two capacitors, in the second circuit below, $C_1$ and $(C_{eq})_{23}$ are in series so their equivalent is given by
$$(C_{eq})_{123}=\left[ \dfrac{1}{C_1}+\dfrac{1}{(C_{eq})_{23}} \right]^{-1}$$
Plug the known;
$$(C_{eq})_{123}=\left[ \dfrac{1}{10}+\dfrac{1}{30} \right]^{-1}$$
$$(C_{eq})_{123}=\color{red}{\bf 7.5}\;\rm \mu F$$
