Answer
$C_{eq} = 41~\mu F$
Work Step by Step
We can find the equivalent capacitance of $4C$ and $6C$ which are in series:
$\frac{1}{C_{46}} = \frac{1}{4C}+\frac{1}{6C}$
$\frac{1}{C_{46}} = \frac{3}{12C}+\frac{2}{12C}$
$C_{46} = \frac{12C}{5}$
The equivalent capacitance of these two capacitors and $2C$ is $C_{246} = 2C+\frac{12C}{5} = \frac{22C}{5}$
We can find the equivalent capacitance of the four capacitors:
$\frac{1}{C_{eq}} = \frac{1}{C}+\frac{1}{22C/5}$
$\frac{1}{C_{eq}} = \frac{1}{C}+\frac{5}{22C}$
$\frac{1}{C_{eq}} = \frac{22}{22C}+\frac{5}{22C}$
$C_{eq} = \frac{22C}{27}$
$C_{eq} = \frac{(22)(50~\mu F)}{27}$
$C_{eq} = 41~\mu F$
