A series circuit has R = 80 Ω, L = 0.32 H and C = 50 µF…

Step 1. The angular frequency. $\omega = 2\pi f = 2\pi \times 50\ \text{Hz} = 314\ \text{rad/s}$.

Step 2. The reactances. $X_L = \omega L = 314\ \text{rad/s} \times 0.32\ \text{H} = 100.5\ \Omega$. $X_C = \dfrac{1}{\omega C} = \dfrac{1}{314\ \text{rad/s} \times 50 \times 10^{-6}\ \text{F}} = 63.7\ \Omega$.

Step 3. The impedance. $X_L - X_C = 36.8\ \Omega$. So $Z = \sqrt{(80\ \Omega)^2 + (36.8\ \Omega)^2} = 88.1\ \Omega$.

Step 4. The current. $I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{230\ \text{V}}{88.1\ \Omega} = 2.61$ A.

Step 5. The phase. $\tan\phi = \dfrac{36.8\ \Omega}{80\ \Omega} = 0.460$, so $\phi = 24.7^\circ$. The circuit is inductive, so the current lags.