Answer
$C_{eq}=3.00\mu F$
Work Step by Step
$C_3$ and $C_5$ are in series so their equivalent capacitance is:
$C_{35}=\frac{C_3C_5}{C_3+C_5}=\frac{4.0\times 10^{-6}\times 4.0\times 10^{-6}}{4.0\times 10^{-6}+4.0\times 10^{-6}}=2.0\times 10^{-6}=2.0\mu F$
Now $C_{35}$ is parallel to $C_2$and $C_4$ so the equivalent capacitance is
$C_{2345}=C_2+C_4+C_{35}=2.0\mu F+2.0\mu F+2.0\mu F=6.0\mu F$
$C_1$ and $C_6$ are also parallel, hence
$C_{16}=C_1+C_6=3.0\mu F+3.0\mu F=6.0\mu F$
Now, $C_{16}$ is in series to $C_{2345}$. So the equivalent capacitance is
$C_{eq}=\frac{C_{2345}C_{16}}{C_{2345}+C_{16}}$
We plug in the known values to obtain:
$C_{eq}=\frac{6.0\times 10^{-6}\times 6.0\times 10^{-6}}{6.0\times 10^{-6}+6.0\times 10^{-6}}=3.00\times 10^{-6}=3.00\mu F$
