In the circuit shown, charge on the $$5\,\mu\text{F}$$ capacitor is:

Taking potential at reference node $$O$$ as $$V_O = 0\text{ V}$$. Let the potential of the upper junction node be $$V_x$$.

Potentials at the outer nodes determined by the batteries:

Left outer node potential: $$V_{\text{left}} = 6\text{ V}$$

Right outer node potential: $$V_{\text{right}} = -6\text{ V}$$

Applying charge conservation at node $$V_x$$: $$C_1(V_x - V_{\text{left}}) + C_2(V_x - V_O) + C_3(V_x - V_{\text{right}}) = 0$$

$$2(V_x - 6) + 5(V_x - 0) + 4(V_x - (-6)) = 0$$

$$2V_x - 12 + 5V_x + 4V_x + 24 = 0 \implies 11V_x + 12 = 0 \implies V_x = -\frac{12}{11}\text{ V}$$

$$q = C_2|V_x - V_O| \implies q = 5 \times \left|-\frac{12}{11} - 0\right| = \frac{60}{11} \approx 5.45\ \mu\text{C}$$