The equivalent capacitance of the system shown in the following circuit is:

For parallel grouping(the loop part), equivalent capacitance can be written as

$C_p=\left(3+3\right) \mathrm{\mu F}=6 \mathrm{\mu F}$.

Now, this part is in series with the left capacitor. So we can write

$C_{eq}=\left(\frac{3\times 6}{3+6}\right) \mathrm{\mu F}=2 \mathrm{\mu F}$