Parallel RLC Resonance bandwidth
Any help would be appreciated.


You will need to calculate L and C to get the required answer. Only with values for L and C can you calculate the magnitude of the impedance when not at resonance. At resonance L and C cancel each other and the input impedance is purely 5 kΩ hence this is the value of the parallel resistor.

So, knowing that Q = 20 and ωO = 10,000 rad/s (at resonance), you can calculate the inductive reactance because Q = R/ωL for a parallel tuned RLC. From the inductive reactance you get L. From L and ωO you get C: -

$$ω_O = \dfrac{1}{\sqrt{LC}}$$

Now you know all the component values for R, L and C.

Next, an impedance magnitude of 3 kohm means that the net reactance is: -

$$\sqrt{5000 Ω^2 - 3000 Ω^2}= 4000Ω$$

In other words a 5 kohm resistor in parallel with a 4 kohm reactance is a magnitude impedance of 3 kohm.

Can you take it from here?