The capacitance between adjacent plates in a three-plate...

Let the charge given to the middle plate be $Q = 1.0 \mu\text{C}$.

By symmetry, the charge $Q$ distributes equally on the two faces of the middle plate. Therefore, the charge on the upper face of the middle plate is $q = \dfrac{Q}{2} = 0.5 \mu\text{C}$.

The upper plate and the middle plate act as a parallel plate capacitor with capacitance $C = 50 \text{ nF}$.

The potential difference $V$ between the upper and middle plates depends on the charge on their facing surfaces.

Using the relation $V = \dfrac{q}{C}$:

$V = \dfrac{0.5 \times 10^{-6}}{50 \times 10^{-9}} = \dfrac{500 \times 10^{-9}}{50 \times 10^{-9}} = 10 \text{ V}$