One plate of a parallel-plate capacitor with a capacitance of 0.1...

Let the charges on the two plates be $Q_1$ and $Q_2$.

The charge appearing on the inner surfaces of the plates is given by $q = \dfrac{|Q_1 - Q_2|}{2}$.

Substituting the given values, $Q_1 = 1 \mu\text{C}$ and $Q_2 = 2 \mu\text{C}$:

$q = \dfrac{|1 - 2|}{2} = 0.5 \mu\text{C}$

The potential difference $V$ between the plates is given by $V = \dfrac{q}{C}$.

Substituting $q = 0.5 \mu\text{C}$ and $C = 0.1 \mu\text{F}$:

$V = \dfrac{0.5}{0.1} = 5 \text{ V}$