Resonance in an LCR circuit occurs when the inductive reactance (XL) and the capacitive reactance (XC) cancel each other out, resulting in a purely resistive circuit. At resonance, the impedance of the circuit is minimum, and the current flowing through the circuit is maximum.

Given parameters:
- Voltage across resistor (R): VR = 100 V
- Resistance (R): 1 kΩ
- Capacitance (C): 2 μF
- Resonant frequency (ω): 200 rad/s

To find the voltage across the inductor (L) at resonance, we'll follow these steps:

1. Calculate the angular frequency of the circuit:
ω = 2πf, where f is the frequency in Hz.
Given ω = 200 rad/s, we can calculate the frequency:
f = ω / 2π = 200 / (2π) ≈ 31.83 Hz

2. Calculate the inductive reactance (XL):
XL = ωL, where L is the inductance in Henrys.
At resonance, XL = XC.
XC = 1/(2πfC), where C is the capacitance in Farads.
Equating the two reactances, we can find the inductance:
ωL = 1/(2πfC)
L = 1/(2πfCω) = 1/(2π * 31.83 * 10^(-3) * 2 * 10^(-6)) ≈ 2.5 H

3. Calculate the current flowing through the circuit at resonance:
V = IZ, where V is the voltage across the circuit, I is the current, and Z is the impedance.
At resonance, the impedance Z = R, as the reactances cancel each other out.
So, I = V / R = 100 / 1000 = 0.1 A

4. Calculate the voltage across the inductor (VL):
At resonance, VL = IXL
VL = 0.1 * 200 ≈ 20 V

Therefore, the voltage across the inductor at resonance is approximately 20 volts. However, the provided correct answer is 250 volts, which seems incorrect based on the given parameters. Please verify the answer or provide additional information to clarify the discrepancy.

At resonance, impedance is purely resistive.