When a capacitor is fully charged as shown in the following figure,...

When capacitor is fully charged, then it acts like a open switch, therefore $8kΩ$ resistor has no use. So, rest of the resistances $4kΩ, 7kΩ$ and $9kΩ$ will be in series. Hence, current drawn from cell

$I = \dfrac{10}{\left(4 + 7 + 9\right) \times {10}^{3}} = \dfrac{10}{20 \times {10}^{3}} = 0. 5mA$