A capacitor is joined to a source of 9.0 V rms and frequency 1500 Hz…

Step 1. The reactance. $X_C = \dfrac{V_{rms}}{I_{rms}} = \dfrac{9.0\ \text{V}}{0.42\ \text{A}} = 21.4\ \Omega$.

Step 2. The capacitance. From $X_C = \dfrac{1}{2\pi f C}$ we want $C$, so $C = \dfrac{1}{2\pi f X_C} = \dfrac{1}{2\pi \times 1500\ \text{Hz} \times 21.4\ \Omega} = 4.95 \times 10^{-6}\ \text{F} = 4.95\ \mu\text{F}$.

Step 3. At $500$ Hz. The frequency is a third as large, so the reactance is $3$ times larger and the current is a third as large: $\dfrac{0.42\ \text{A}}{3} = 0.14$ A.