Let the charges on the $ 3 \mu \mathrm{F} $ and $ 2 \mu \mathrm{F} $ capacitors be $ q $ and $ q' $ respectively. The charge on the lower plate of the $ 4 \mu \mathrm{F} $ capacitor is $ -80 \mu \mathrm{C} $. The lower plate of the $ 4 \mu \mathrm{F} $ capacitor and the upper plates of the $ 2 \mu \mathrm{F} $ and $ 3 \mu \mathrm{F} $ capacitors form an isolated system. Therefore, the net charge on this system must be zero, which can be represented as :

$ q + q' - 80 \mu \mathrm{C} = 0 \quad \text{..........(i)} $

The potentials ($ V = q / C $) across these two capacitors are equal, which gives :

$ \frac{q}{3} = \frac{q'}{2} \quad \text{..........(ii)} $

By substituting $ q' $ from equation (ii) into equation (i), we can solve for $ q $ and find :

$ q = 48 \mu \mathrm{C} $