A series circuit has R = 60 Ω, XL = 145 Ω and XC = 65 Ω…

Step 1. The parts. $V_R = IR = 1.2\ \text{A} \times 60\ \Omega = 72$ V. $V_L = IX_L = 1.2\ \text{A} \times 145\ \Omega = 174$ V. $V_C = IX_C = 1.2\ \text{A} \times 65\ \Omega = 78$ V.

Step 2. The impedance. $X_L - X_C = 80\ \Omega$. So $Z = \sqrt{(60\ \Omega)^2 + (80\ \Omega)^2} = 100\ \Omega$.

Step 3. The source. $V = IZ = 1.2\ \text{A} \times 100\ \Omega = 120$ V.