Answer
$7.33\ uF$
Work Step by Step
in the given figure $C_1 $ and $C_2$ are in series. Their equivalent capacitance is;
$\frac {1}{C_{12}} = \frac{1}{C_1}+\frac{1}{C_2}$
$\frac {1}{C_{12}} = \frac{1}{10\mu F}+\frac{1}{5\mu F}$
$C_{12} = \frac{10\times5}{10+5} = 3.33\mu F$
Also,
$C_{12} $ and $C_3$ are in parallel. So the total equivalent capacitance is;
$C_{equ} = C_{12} + C_{3} = 3.33\mu F + 4\mu F=7.33 \mu F$
