The capacitance between the adjacent plates depicted in the figure...

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

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

The total charge on the system of plates is:
$Q_{\text{net}} = Q_1 + Q_2 + Q_3 = 0 + 1.0 \mu\text{C} + 0 = 1.0 \mu\text{C}$

For a system of parallel conducting plates, the electric field inside the bulk of any conducting plate must be zero. By applying Gauss's law or the principle of superposition of electric fields due to charged sheets, it can be shown that the charge appearing on the outermost surfaces of the system is always equal and is exactly half of the total net charge of the system.

Let $q_{\text{out}}$ be the charge on the outer surface of the upper plate.
$q_{\text{out}} = \dfrac{Q_{\text{net}}}{2}$

Substituting the value of $Q_{\text{net}}$:
$q_{\text{out}} = \dfrac{1.0 \mu\text{C}}{2} = 0.50 \mu\text{C}$

The capacitance value given ($50 \text{ nF}$) is extra information and is not required to find the charge distribution on the outermost surfaces.