In an RLC series circuit, the rms potential difference...

In an RLC series circuit, the rms potential difference provided by the source is V =160 V, and the frequency is f = 250/pi Hz. Given that L = 0.6 H, C = 45 µF, and Vr = 20 V, find:
I(rms)
R
Vl(rms)
Vc(rms)

I keep finding I = 0.406 A...... what am i doing wrong?

Irms is 160/Z where Z = R+jwL-j/(wC).
wL= (250*2]*0.6 = 300.
1/(wC) = 44.4
One method of calculation is by assuming various values of R and identify for what R, voltage is 20V.
I found that R = 32.3 results in 20V across R.
If the jwL-j/(wC) cancels, you have resonance. In that case, VR = 160V.
But since VR is 20V, the circuit is not in resonance. The resonance freq = 1/[2.pi.sqrt(LC)], and is 96.22/pi. Since operating frequency here is 250/pi, you are at a frequency where wL dominates. Neglecting C, the Z = sqrt(R+(wL)^2).

Current is about 620mA and R is 32.3. Product gives 20V.
VL is 186.7 and
Vc =27.7V, from simulation.
From this, Z= 32.3+j(300-44)=32.3+j256=258, and current is 620mA.
The resonance frequency is 30.6Hz and one gets VL =Vc as about 569V at this frequency. the current is 4.92 amps.
I used "circuit maker" simulation tool. [free download].

** Zʟ = 2 π fC = 2 π * 250/pi * 0.6 = 300 Ω
Xc =1/( 2 π fC ) = 1/ ( 2 π * 250/pi * 45 ^-6) = 44.4444 Ω

Z of L C alone in series = 300Ω - 44.4444 Ω= 255.5556 Ω
V across L C in series = 160V - 20V = 140V . This OK as Vresistor and I in phase.

I L C in series
= V across L C in series / Z of L C alone in series
= 140V / 255.5556Ω = 0.54783 A = Circuit current = Part Answer
R = V resistor / I resistor
= 20V / 0.54783A = 36.5079 Ω = Resistor Value = Part Answer
Vʟ = I * Xʟ = 0.54783A * 300 Ω = 154.349V = Vʟ =Part Answer
Vc = I * Xc = 0.54783A * 44.4444 = 24.348V = Vc = Part Answer

Clarification of any point will be provided, if requested.