1. Given data

   $L=0.01\,\text{H},\quad R=10\,\Omega,\quad C=1\,\mu\text{F}=10^{-6}\,\text{F}$ $V_m=50\,\text{V}$

   The circuit is driven at a frequency 60% lower than the resonant frequency.

2. Resonant angular frequency

   For a series LCR circuit, $\omega_0=\frac{1}{\sqrt{LC}}$

   Compute: $LC=0.01\times 10^{-6}=10^{-8}$ $\sqrt{LC}=10^{-4}$ $\omega_0=\frac{1}{10^{-4}}=10^4\,\text{rad/s}$

3. Operating frequency

   “60% lower than resonant frequency” means $\omega=0.4\,\omega_0=0.4\times 10^4=4000\,\text{rad/s}$

4. Reactances at this frequency

   Inductive reactance: $X_L=\omega L=4000\times 0.01=40\,\Omega$

   Capacitive reactance: $X_C=\frac{1}{\omega C}=\frac{1}{4000\times 10^{-6}}=\frac{1}{4\times 10^{-3}}=250\,\Omega$

   Net reactance: $X=X_L-X_C=40-250=-210\,\Omega$

5. Impedance of the series circuit

   $Z=\sqrt{R^2+(X_L-X_C)^2}$ $Z=\sqrt{10^2+210^2}$ $Z=\sqrt{100+44100}$ $Z=\sqrt{44200}\approx 210.24\,\Omega$

6. Current amplitude

   $I_m=\frac{V_m}{Z}=\frac{50}{210.24}\approx 0.238\,\text{A}$

   $I_m\approx 238\,\text{mA}$

7. Option check

   • A: $466\,\text{mA}$ ❌
   • B: $312\,\text{mA}$ ❌
   • C: $238\,\text{mA}$ ✅
   • D: $196\,\text{mA}$ ❌

Therefore, the correct option is C.