A

Step 1: The force exerted by the electric field on the capacitor plates can be derived from the electric field between the plates, which is given by:
$$ E = \frac{V}{d} $$
where $V$ is the voltage and $d$ is the separation between the plates.
Step 2: The force $F$ on one plate is calculated using:
$$ F = qE $$
where $q$ is the charge on the plate given by
$$ q = C \cdot V $$
Step 3: The capacitance $C$ of a parallel plate capacitor can be expressed as:
$$ C = \frac{\varepsilon_0 A}{d} $$
where $A$ is the area of the plates. Hence,
$$ q = \frac{\varepsilon_0 A V}{d} $$
Step 4: We substitute to find the force:
$$ F = \left( \frac{\varepsilon_0 A V}{d} \right) \left( \frac{V}{d} \right) = \frac{\varepsilon_0 A V^2}{d^2} $$
Step 5: This force causes an extension in the spring ($x$) according to Hooke's law:
$$ F = K_s x $$
where $K_s$ is the spring constant. Equating the two forces, we get:
$$ K_s x = \frac{\varepsilon_0 A V^2}{d^2} $$
Step 6: From this equation, we can solve for the extension $x$:
$$ x = \frac{\varepsilon_0 A V^2}{K_s d^2} $$
Therefore, the extension in the spring in equilibrium is given by:
$$ x = \frac{\varepsilon_0 A V^2}{K_s d^2} $$
Thus, the correct answer is A.