Answer
$2.56\mu F$
Work Step by Step
We know that
$\frac{1}{C_{12}}=\frac{1}{12.0\mu F}+\frac{1}{8.35\mu F}$
$\implies C_{12}=4.924\mu F$
and $C_{123}=C_{12}+C_{3}=4.924\mu F+4.25\mu F=9.174\mu F$
Now $\frac{1}{C_{1234}}=\frac{1}{C_{123}}+\frac{1}{C_4}$
$\implies \frac{1}{C_{1234}}=\frac{1}{9.174\mu F}+\frac{1}{C}$
$\implies C_{1234}=\frac{(9.174\mu F)(C)}{C+9.174\mu F}$
Now $C_{eq}=C_{1234}+C_5$
We plug in the known values to obtain:
$9.22\mu F=(\frac{(9.174\mu F)(C)}{C+9.174\mu F})+7.22\mu F$
This simplifies to:
$C=2.56\mu F$
