In the following circuit the resultant capacitance between A and B is $1 \mu F$.  Then value of C is 

12$\mu F$ and 6$\mu F$  are in series and again are in parallel with4$\mu F$ therefore resultant of these three will be 
$=\frac{12\times 6}{12+6}+4=4+4=8\mu F$
This  equivalent system is in series with 1$\mu F$ 
Its equivalent capacitance$=\frac{8\times 1}{8+1}=\frac{8}{9}\mu F\to \left(1\right)$ 
Equivalent of 8$\mu F$ , 2$\mu F$  and 2$\mu F$ 
$=\frac{4\times 8}{4+8}=\frac{8}{3}\mu F\to \left(2\right)$
1 and 2 are in parallel and are in series with C
$$\begin{gathered} \therefore \frac{8}{9}+\frac{8}{3}=\frac{32}{9} \\ {C}_{eq}=1=\frac{\frac{32}{9}\times C}{\frac{32}{9}+C} \\ \Rightarrow C=\frac{32}{23}\mu F \end{gathered}$$