A series circuit has R = 30 Ω, L = 0.20 H and C = 20 µF…

Step 1. At $40$ Hz. $\omega = 2\pi \times 40 = 251\ \text{rad/s}$. $X_L = 251\ \text{rad/s} \times 0.20\ \text{H} = 50.3\ \Omega$. $X_C = \dfrac{1}{251\ \text{rad/s} \times 20 \times 10^{-6}\ \text{F}} = 199\ \Omega$. The net reactance is $-149\ \Omega$. So $\tan\phi = \dfrac{-149\ \Omega}{30\ \Omega} = -4.96$ and $\phi = -78.6^\circ$. The capacitor wins, and the current leads.

Step 2. At $200$ Hz. $\omega = 2\pi \times 200 = 1257\ \text{rad/s}$. $X_L = 1257\ \text{rad/s} \times 0.20\ \text{H} = 251\ \Omega$. $X_C = \dfrac{1}{1257\ \text{rad/s} \times 20 \times 10^{-6}\ \text{F}} = 39.8\ \Omega$. The net reactance is $+212\ \Omega$. So $\tan\phi = \dfrac{212\ \Omega}{30\ \Omega} = 7.05$ and $\phi = 81.9^\circ$. The coil wins, and the current lags.