Assuming potential x and y at the top junctions A and B respectively and zero potential at lower junctions.

At the junction A applying KCL,

i₁ + i₂ + i₃ = 0

$${{x - 72} \over 6} + {x \over 4} + {{x - y} \over 2} = 0$$

$$ \Rightarrow $$ $${{2x - 144 + 6x - 6y + 3x} \over {12}} = 0$$

$$ \Rightarrow $$ 11x - 6y = 144 ......(1)

At the junction B applying KCL,

i₄ + i₅ = 0

$${{y - x} \over 2} + {y \over {10}} = 0$$

$$ \Rightarrow $$ $${{5y - 5x + y} \over {10}}$$ = 0

$$ \Rightarrow $$ 6y = 5x ......(2)

Solving (1) and (2)

11x - 5x = 144

$$ \Rightarrow $$ x = 24 V

$$ \therefore $$ y = 20 V

The charge on the capacitor, q = CV

$$ \Rightarrow $$ q = 10$$\mu $$F × 20 = 200 $$\mu $$C