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

$I=\frac{10}{\left(4+7+9\right)\times {10}^3}=\frac{10}{20\times {10}^3}=0.5 \mathrm{mA}$