Separation are d, 2d and 3d , then the equivalent capacitance across A and area of each plate A and the successive B is

Plates can be rearranged as shown in figure. Plates 1 and 2 and plates 3 and 4 form two capacitors which are in series between A and B. Plates 2 and 3 do not form any capacitor as they are at same potential.

So $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{\left(\frac{{\epsilon }_{0}A}{d}\right)\times \left(\frac{{\epsilon }_{0}A}{3d}\right)}{\frac{{\epsilon }_{0}A}{d}+\frac{{\epsilon }_{0}A}{3d}}$

$=\frac{{\epsilon }_{0}A}{4d}$