1. Model the geometry

A flat plate of total area $A$ faces a stair-shaped conductor.
Each stair has area $\dfrac{A}{3}$, so the arrangement can be treated as three parallel capacitors, each occupying area $\dfrac{A}{3}$ but with different plate separations.

Since the stair height is $d$, the three levels are at distances:

$d,\; 2d,\; 3d$

from the flat plate.

2. Capacitance of each part

For a parallel plate capacitor,

$C = \frac{\varepsilon_0 A}{\text{separation}}$

For each stair region of area $\dfrac{A}{3}$:

• First region: $C_1 = \frac{\varepsilon_0 (A/3)}{d} = \frac{\varepsilon_0 A}{3d}$

• Second region: $C_2 = \frac{\varepsilon_0 (A/3)}{2d} = \frac{\varepsilon_0 A}{6d}$

• Third region: $C_3 = \frac{\varepsilon_0 (A/3)}{3d} = \frac{\varepsilon_0 A}{9d}$

3. Combine them

These three capacitors are in parallel because all parts of the stair conductor are connected together and face the same flat plate.

So,

$C_{\text{eq}} = C_1 + C_2 + C_3$

$C_{\text{eq}} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d}$

Take common factor $\dfrac{\varepsilon_0 A}{d}$:

$C_{\text{eq}} = \frac{\varepsilon_0 A}{d}\left(\frac13 + \frac16 + \frac19\right)$

Now,

$\frac13 + \frac16 + \frac19 = \frac{6+3+2}{18} = \frac{11}{18}$

Hence,

$C_{\text{eq}} = \frac{11\varepsilon_0 A}{18d}$

4. Match with the options

$\boxed{C = \frac{11\varepsilon_0 A}{18d}}$

This corresponds to Option D.