At resonance in a series LCR circuit

The inductive and capacitive reactances cancel: $X_L = X_C$ so the impedance becomes purely resistive: $Z = R$

Current at resonance

Given: $V = 250\ \text{V}, \quad R = 5\ \Omega$

Therefore, $I = \frac{V}{R} = \frac{250}{5} = 50\ \text{A}$

Power dissipated at resonance

In a series LCR circuit, only the resistor dissipates power. Thus, $P = I^2 R$

Substituting: $P = (50)^2 \cdot 5 = 2500 \cdot 5 = 12500\ \text{W}$

Match with the required form

The question asks for: $\underline{\hspace{2cm}} \times 10^2\ \text{W}$

Since $12500 = 125 \times 10^2$

the required integer is: $125$

Comparison with stored answer

Derived answer = $125$

Stored correct answer = $125$

Hence, they agree.