Three parallel plates are arranged one above the other, with a...

Let the three parallel plates be $P_1$ (upper), $P_2$ (middle), and $P_3$ (lower).

The total charge given to the system is $Q = 1.0 \mu\text{C}$ on the upper plate, while the middle and lower plates are uncharged ($Q_2 = 0$, $Q_3 = 0$).

For a system of parallel conducting plates, the charge appearing on the outermost surfaces is half of the total charge of the system.
Charge on the outermost surfaces $= \dfrac{Q_{total}}{2} = \dfrac{1.0 \mu\text{C} + 0 + 0}{2} = 0.5 \mu\text{C}$.

Thus, the top surface of $P_1$ has a charge of $0.5 \mu\text{C}$.
Since the total charge on $P_1$ is $1.0 \mu\text{C}$, the charge on its bottom surface is:
$q_{1, \text{bottom}} = 1.0 \mu\text{C} - 0.5 \mu\text{C} = 0.5 \mu\text{C}$.

Facing surfaces of adjacent conducting plates always carry equal and opposite charges.
Charge on the top surface of $P_2 = -0.5 \mu\text{C}$.
Since $P_2$ is isolated and has zero net charge, the charge on its bottom surface is:
$q_{2, \text{bottom}} = 0 - (-0.5 \mu\text{C}) = +0.5 \mu\text{C}$.

Similarly, the charge on the top surface of $P_3$ will be $-0.5 \mu\text{C}$.

The middle and lower plates form a capacitor with capacitance $C = 50 \text{ nF}$ and charge $q = 0.5 \mu\text{C} = 500 \text{ nC}$ on their facing surfaces.

The potential difference $V$ between the middle and lower plates is:
$V = \dfrac{q}{C} = \dfrac{500 \times 10^{-9} \text{ C}}{50 \times 10^{-9} \text{ F}} = 10 \text{ V}$.