In the given fig. the resultant capacity will be-
 

First n capacitor are connected in series so, their resulted capacity  $C_1=C/n$

 second n capacitor are connected in parallel so, their resultant capacitance $C_2=nC$ . Now  $C_{1}and{C}_2$ are in series, so resultant capacity 
 $$\begin{gathered} \frac{1}{C^{\text{'}}}=\frac{1}{C_1}+\frac{1}{C_2} \\ \Rightarrow \frac{1}{C^{\text{'}}}=\frac{n}{C}+\frac{1}{nC}=\frac{n^2+1}{nC} \end{gathered}$$

$\Rightarrow C^{\text{'}}=\frac{n}{n^2+1}C$
   
NOTE: In series charge remains same on each capacitor and voltage remains same on each capacitor in parallel.