Equivalent R for Bandwidth of non-ideal parallel RLC circuit
To determine the impedance offered by this 2nd-order network (two energy-storing elements with independent state variables), you have two options: the brute-force approach or the smarter one using the fast analytical circuits techniques or FACTs.

The brute-force is simply to write \$Z(s)=R_C||\frac{1}{sC_1}||(r_L+sL_2)\$. You can expand this expression and factor it under a nicely-looking canonical form obeying \$Z(s)=R_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$. Good luck in doing that because a) you can make mistakes while developing the expression b) you will end-up with numerous non-ordered terms and c) you will have to inject more energy to rearrange the expression into the canonical form I gave. Why doing this at the end? Because you must highlight what will let you fulfill your design goal: a certain quality factor \$Q\$, an ohmic peak and a tuning frequency \$\omega_0\$. The FACTs will lead you straight to this final point, without writing a single line of algebra, just drawings!

To determine an impedance, we install a test generator - the stimulus -  \$I_T\$ generating a response across its terminals - the response - \$V_T\$. The impedance we want is \$Z(s)=\frac{V_T}{I_T}\$.

We first start with \$s=0\$: short the inductor and open the capacitor as shown below. Then, determine by inspection the resistance \$R_0\$ in this mode. Without an equation, you see that this is \$R_C||r_L\$. Then, determine the two time constants \$\tau_1\$ and \$\tau_2\$ as shown in the below sketches in which the excitation has been reduced to 0 A: what resistance do you "see" from the selected energy-storing elements while the second one is in its dc state (shorted inductor or open capacitor). Once you have the time constants, you form \$b_1=\tau_1+\tau_2\$.



The last time constant \$\tau_{12}\$ is determined when the capacitor is replaced by a short circuit and you "see" the resistance from \$L_1\$'s connections: \$\tau_{12}=\frac{L_2}{r_L}\$. We have \$b_2=\tau_1\tau_{12}\$ and we can form the denominator as \$D(s)=1+s(\tau_1+\tau_2)+s^2\tau_1\tau_{12}\$ and put it under the canonical form \$D(s)=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$. The zero is obtained by realizing that the series combination of \$r_L\$ and \$L_2\$ forms a transformed short circuit nulling the response. The zero is thus defined as \$\omega_z=\frac{r_L}{L_2}\$ and the complete impedance expression is \$Z(s)=R_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$.

However, this expression indicates that \$r_L\$ dominates at low frequency. What we want - and what you need - is the peak at the resonance. Factor \$\frac{s}{\omega_0Q}\$ in the denominator and \$\frac{s}{\omega_z}\$ in the numerator. You find that the peak is defined as \$\frac{R_0Q\omega_0}{\omega_z}\$. You know the peak, you know \$Q\$, I leave the rest to you.



You have seen how the FACTs could get you straight to the result via a few sketches, without a line of algebra. This is truly a tool you want to master if you have to determine transfer functions of any kind.