Four identical large metal plates, each of area A , are arranged...

Let the four plates be numbered 1, 2, 3, and 4 from top to bottom.

Three individual capacitors are formed by the adjacent pairs of plates: between (1, 2), between (2, 3), and between (3, 4).

The capacitance of each individual capacitor is $C = \dfrac{\varepsilon_0 A}{d}$.

Let the first and third plates be connected to terminal X, and the second and fourth plates be connected to terminal Y.

The first capacitor (plates 1 and 2) is connected between X and Y.
The second capacitor (plates 2 and 3) is connected between Y and X.
The third capacitor (plates 3 and 4) is connected between X and Y.

Since all three capacitors are connected directly across the same two terminals X and Y, they are in a parallel combination.

The equivalent capacitance $C_{\text{eq}}$ of three capacitors in parallel is:
$C_{\text{eq}} = C + C + C = 3C$

Substituting the value of $C$:
$C_{\text{eq}} = \dfrac{3\varepsilon_0 A}{d}$

Answer: $\dfrac{3\varepsilon_0 A}{d}$