Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance \(C\) if the equivalent capacitance between point \(A\) and \(B\) is to be \(1\,\mu F\) is 

\(6 \,\mu \mathrm{F}\) and \(12\, \mu \mathrm{F}\) are in series, their equivalent \(=4\, \mu \mathrm{F}\)

Now \(4 \,\mu F(2 \text { and } 2\, \mu F)\) and \(8\, \mu F\) in series

\(=\frac{3}{8}\, \mu F\)

And \(4\, \mu \mathrm{F}(12 \& 6\, \mu \mathrm{F})\) and \(4\, \mu \mathrm{F}\) in parallel \(=4+4=8\, \mu \mathrm{F}\)

\(8\, \mu \mathrm{F}\) in series with \(1\, \mu \mathrm{F}=\frac{1}{8}+1 \Rightarrow \frac{8}{9}\, \mu \mathrm{F}\)

Now \(C_{e q}=\frac{8}{9}+\frac{8}{3}=\frac{32}{9}\)

\(C_{eq}\) of circuit \(=\frac{32}{9}\)

With \(C-\frac{1}{C_{e q}}=\frac{1}{C}+\frac{9}{32}=1 \Rightarrow C=\frac{32}{23}\)