Answer
$3.157\mu F$
Work Step by Step
In the given circuit $C_1 $and $C_2$ are in parallel so their equivalent capacitance is $C_{12} = C_1+C_2 = 10\mu F + 5\mu F=15\mu F$
Next, we know that $C_{12}$ and $C_{3}$ are in series so the total equivalent capacitance is;
$\frac{1}{C_{equ}}=\frac{1}{C_{12}}+\frac{1}{C_3}$
$\frac{1}{C_{equ}}=\frac{1}{15}+\frac{1}{5}$
$C_{equ} = \frac{15\times 5}{15+5}= 3.157\mu F$