At resonance in a series LCR circuit

The inductive and capacitive reactances cancel: $X_L = X_C$ So the impedance becomes purely resistive: $Z = R = 8\,\Omega$

Given peak voltage

Peak voltage: $V_0 = 250\,\text{V}$

RMS voltage is: $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{250}{\sqrt{2}}\,\text{V}$

Power dissipated at resonance

Since the circuit is purely resistive at resonance, $P = \frac{V_{\text{rms}}^2}{R}$

Substitute values: $P = \frac{\left(\frac{250}{\sqrt{2}}\right)^2}{8}$

$P = \frac{250^2/2}{8} = \frac{62500}{16}$

$P = 3906.25\,\text{W}$

Convert to kW

$P = 3.90625\,\text{kW}$

Therefore, $x \approx 4$ to the nearest integer.

Comparison with stored answer

Derived answer: $4$

Stored correct answer: $4$

Hence, the answers agree.