Step 1: Lay out the circuit.
A resistor $R$ and a capacitor $C$ are in series across an AC source. At angular frequency $\omega$ the current is $I$. When the frequency drops to $\omega/3$ (voltage kept the same), the current falls to $I/2$. We want $\frac{X_c}{R}$ at the first frequency.
Step 2: Recall reactance and impedance.
The capacitive reactance is $X_c = \frac{1}{\omega C}$, and the total opposition is $Z = \sqrt{R^2 + X_c^2}$, with current $I = \frac{V}{Z}$.
Step 3: See how reactance changes.
When the frequency becomes one third, the reactance triples: $X_c' = 3X_c$. So the new impedance is $Z' = \sqrt{R^2 + 9X_c^2}$.
Step 4: Use the halved current.
Half the current means double the impedance, so $Z' = 2Z$, that is $\sqrt{R^2 + 9X_c^2} = 2\sqrt{R^2 + X_c^2}$.
Step 5: Square and simplify.
Squaring: $R^2 + 9X_c^2 = 4R^2 + 4X_c^2$, which gives $5X_c^2 = 3R^2$, so $\frac{X_c^2}{R^2} = \frac{3}{5} = 0.6$.
Step 6: Take the root.
Therefore $\frac{X_c}{R} = \sqrt{0.6}$.
\[ \boxed{\frac{X_c}{R} = \sqrt{0.6}} \]