As show in the figure, in steady state, the charge stored in the capacitor is……. × 10^–6C.

In steady state condition, There will be no current passing through the capacitor.

So the potential difference across the capacitor will be equal to the potential drop across the 100 Ohm resistor.

$\begin{array}{l}\mathrm{q}={\mathrm{C}}_{100\mathrm{\Omega }}\\ =\left(1.1\times {10}^{-6}\right)\left(\frac{10}{\mathrm{R}+\mathrm{r}}\mathrm{R}\right)\\ =1.1\times {10}^{-6}\left(\frac{10}{110}\times 100\right)\\ =10\mathrm{\mu C}\end{array}$