A practical approach is possible if you have a physical circuit at hand; you send in electrical energy ( 1 V ac and 1 mA for example ) and sweep up the frequency spectrum. Then you will find one or two resonance frequencies at which the Q is higher than the general spectrum of a pure resistive element. If the voltage goes up to 3 V at the same injected current it means that the Q is 3 at this resonance frequency. If you change the diameter of the coil or change the ferrite core material, this will change the Q. If you add resistance this will reduce the Q.

The quality factor Q is defined for reactive elements to show how far they are unlossy especially when they are built in a resonance circuit. If the reactance of the reactive element is X and its equivalent series resistance is rs, then its Q= X/ rs.  If the equivalent Resistance is in parallel and =rp, then the Q= rp/ X.

In case of series resonance circuit which is composed of LCR in series, at resonance 

XC= XL and the circuit Q= XL/ R=XC/R with R is the sum of all the resistances in series XL is the total inductive reactance and XC is the total capacitive reactance at reonance.,i.e., w=wo.

Your input circuit is equivalent to a series resonance circuit with a total equivalent series Resistance R= Rs+ gm LS/ Ct, where the second term gm LS/Ct is the equivalent input resistance between the gate and ground. Ct is equivalent total capacitance in series with Lg. Simple circuit analysis will show that your capacitance Ct is  equal to Cgs. The total inductance will be = Lg+ Ls. So,according to the formula of Q,

Q= XC/R = 1/  [Rs+ gm LS/ Ct] wo Ct as XC= 1/wo Ct.

For the condition of matching Rs= gm LS/ Ct, then we get Q= 1/ 2 Rs Ct wo,

Ct =Cgs . The resonance frequency WO= sq root of 1/ [Lg+ Ls] Cgs. Then you can calculate it knowing the parameters.

This may shed some light on the answer of your question.