A 120-V rms voltage at 1000 Hz is applied to a 2.00-mH inductor, a 4.00-μF capacitor and a resistor.

If the rms value of the current circuit is .400A, what is the value of the resistor?

calculate the reactance of the L and C. Then the impedance of the entire three, see equations below, leaving R as a variable. Then total Z is 120/0.4 = 300 ohms.

Capacitive Reactance Xc = 1/(2πfC) = 1/ωC
Inductive Reactance Xʟ = 2πfL = ωL
Impedance Z = √(R² + X²)
where X = Xʟ – Xc
ω = 2πf
Phase angle θ = arctan (X/R)
Xc, Xʟ, Z are in Ω, f is in Hz
C in farads, L in Henrys

X = 2πfL – (1/(2πfC))
X = 2π1000•2e-3 – (1/(2π1000•4e-6))
300 = √(R² + X²)
R² + X² = 90000
R² = 90000 – 2π1000•2e-3 – (1/(2π1000•4e-6))

the rest is just math.

resistance=sqrt(R^2 + (Xc-Xl)^2)
where R=resistance to be found
Xc=capacitive reactance=1/ωC=1 /(1000 ×2 × 3.14 × 4 × 10^-6)=39.8 ohm (ω=2πf,f=1000 Hz)
Xl=inductive reactance= ωL= 1000 × 2 ×10^-3 × 2 × 3.14 =12.56 ohm
resistance=sqrt(R^2 + (39.8-12.56)^2)
voltage/current=sqrt(R^2 + (27.24)^2)
120/0.4=sqrt(R^2 + 742.01)
on solving,R=298.76 ohm

XL = (6.28)*(1,000)*(.002) = 12.56 Ohm
XC = 1/ [(6.28)*(1,000)*(.000004)] = 39.81 Ohm
net reactance = (12.56 - 39.81)Ohm = -27.25 Ohm (negative sign indicates that the net reactance is capacitive)
Z = 120V/.4A = 300 Ohm
R = (300)*[cos(arc sin -27.25/300)] = 298.76 Ohms