Six plates of equal area A and plate separation as shown (figure) are arranged. The equivalent capacitance between, A and B is

Given plates can be rearranged as shown: 

$C_1=\frac{{\epsilon }_{0}A}{d}; C_2=\frac{{\epsilon }_{0}A}{2d}$

$C_3=\frac{{\epsilon }_{0}A}{3d}; C_4=\frac{{\epsilon }_{0}A}{2d}; C_5=\frac{{\epsilon }_{0}A}{d}$

$C_1 and C_2$

$\frac{\frac{{\epsilon }_{0}A}{d}\times \frac{{\epsilon }_{0}A}{2d}}{\frac{{\epsilon }_{0}A}{d}+\frac{{\epsilon }_{0}A}{2d}}=\frac{{\epsilon }_{0}A}{3d}$

Effective capacitance of 

$C_4 and C_5=\frac{{\epsilon }_{0}A}{3d}$

$C_{AB}=\frac{{\epsilon }_{0}A}{3d}+\frac{{\epsilon }_{0}A}{3d}+\frac{{\epsilon }_{0}A}{3d}=\frac{{\epsilon }_{0}A}{d}$