#### Answer

$C = 25~\mu F$

#### Work Step by Step

We can find the equivalent capacitance $C_s$ of the two capacitors in series:
$\frac{1}{C} = \frac{1}{20\times 10^{-6}~F}+\frac{1}{30\times 10^{-6}~F}$
$\frac{1}{C} = \frac{3}{60\times 10^{-6}~F}+\frac{2}{60\times 10^{-6}~F}$
$\frac{1}{C} = \frac{5}{60\times 10^{-6}~F}$
$C = \frac{60\times 10^{-6}~F}{5}$
$C = 12~\mu F$
We can find the equivalent capacitance $C$:
$C = 12~\mu F + 13~\mu F$
$C = 25~\mu F$