A capacitor consists of a flat plate of area A and another plate...

The given capacitor can be considered as a parallel combination of three smaller capacitors.

The total area of the flat plate is $A$ and its total length is $3a$. Therefore, the area of each stair section of width $a$ is $\dfrac{A}{3}$.

The distances between the plates for the three sections are $d$, $d+b$, and $d+2b$.

The capacitances of the three sections are:
$C_1 = \dfrac{\varepsilon_0 (A/3)}{d}$
$C_2 = \dfrac{\varepsilon_0 (A/3)}{d+b}$
$C_3 = \dfrac{\varepsilon_0 (A/3)}{d+2b}$

Since the three capacitors are connected in parallel, the equivalent capacitance is:
$C_{eq} = C_1 + C_2 + C_3$

C_{eq} = \dfrac{\varepsilon_0 A}{3} \left(\dfrac{1}{d} + \dfrac{1}{d+b} + \dfrac{1}{d+2b} \right)

$C_{eq} =$\dfrac{\varepsilon_0 A}{3} \left[ \dfrac{(d+b)(d+2b) + d(d+2b) + d(d+b)}{d(d+b)(d+2b)} \right]$<br><br>$C_{eq} = \dfrac{\varepsilon_0 A}{3} \left[ \dfrac{(d^2 + 3bd + 2b^2) + (d^2 + 2bd) + (d^2 + bd)}{d(d+b)(d+2b)} \right]<br><br>C_{eq} = $\dfrac{\varepsilon_0 A (3d^2 + 6bd + 2b^2)}{3d(d+b)(d+2b)}$