A circuit consists of a battery of EMF 18 V and internal resistance...

In steady state, the capacitor is fully charged and acts as an open circuit, so no current flows through the branch containing $R_3$ and $C$.

The potential difference across the capacitor is given by $V_C = \dfrac{Q}{C} = \dfrac{18\ \mu\text{C}}{3\ \mu\text{F}} = 6\text{ V}$.

Since there is no current through $R_3$, the voltage drop across $R_3$ is zero. Therefore, the potential difference across the entire parallel combination is equal to the voltage across the capacitor, which is $6\text{ V}$.

This same potential difference of $6\text{ V}$ appears across the first branch containing $R_2$. The current through $R_2$ is $I = \dfrac{V}{R_2} = \dfrac{6}{2} = 3\text{ A}$.

Since no current flows through the capacitor branch, this $3\text{ A}$ is the total current supplied by the battery. Applying Kirchhoff's voltage law to the outer loop:
$E - I r - I R_1 - V_{\text{parallel}} = 0$
$18 - 3(1) - 3 R_1 - 6 = 0$
$12 - 3 - 3 R_1 = 0$
$9 = 3 R_1$
$R_1 = 3\ \Omega$.

Answer: $3\ \Omega$