Answer
$C_{eq} = 42~\mu F$
Work Step by Step
We can find the equivalent capacitance of $2C$ and $6C$ which are in series:
$\frac{1}{C_{26}} = \frac{1}{2C}+\frac{1}{6C}$
$\frac{1}{C_{26}} = \frac{3}{6C}+\frac{1}{6C}$
$C_{26} = \frac{3C}{2}$
The equivalent capacitance of these two capacitors and $4C$ is $C_{246} = 4C+\frac{3C}{2} = \frac{11C}{2}$
We can find the equivalent capacitance of the four capacitors:
$\frac{1}{C_{eq}} = \frac{1}{C}+\frac{1}{11C/2}$
$\frac{1}{C_{eq}} = \frac{1}{C}+\frac{2}{11C}$
$\frac{1}{C_{eq}} = \frac{11}{11C}+\frac{2}{11C}$
$C_{eq} = \frac{11C}{13}$
$C_{eq} = \frac{(11)(50~\mu F)}{13}$
$C_{eq} = 42~\mu F$
