For the arrangement of plates depicted in the figure, the...

Let the three parallel plates be denoted as 1 (upper), 2 (middle), and 3 (lower).

The total charges on the isolated plates are given as:
$Q_1 = 1.0 \mu\text{C}$
$Q_2 = 0$
$Q_3 = 0$

For a system of parallel conducting plates, the charge appearing on the outermost surfaces is equal and is given by half the sum of the total charges on all plates:
$q_{\text{out}} = \dfrac{Q_1 + Q_2 + Q_3}{2} = \dfrac{1.0 \mu\text{C} + 0 + 0}{2} = 0.5 \mu\text{C}$

The charge on the upper surface of plate 1 is $q_{1a} = 0.5 \mu\text{C}$.
The charge on the lower surface of plate 1 is $q_{1b} = Q_1 - q_{1a} = 1.0 \mu\text{C} - 0.5 \mu\text{C} = 0.5 \mu\text{C}$.

The charge on the upper surface of plate 2 (facing plate 1) will be equal and opposite to $q_{1b}$:
$q_{2a} = -q_{1b} = -0.5 \mu\text{C}$

The upper and middle plates form a capacitor with charge $q = 0.5 \mu\text{C}$ on its inner surfaces.
Given the capacitance between adjacent plates is $C = 50 \text{ nF}$.

The potential difference $V$ between the upper and middle plates is:
$V = \dfrac{q}{C} = \dfrac{0.5 \times 10^{-6} \text{ C}}{50 \times 10^{-9} \text{ F}} = \dfrac{500 \times 10^{-9}}{50 \times 10^{-9}} \text{ V} = 10 \text{ V}$