How do I fit a curve into non linear data?
Here is a simple solution that you can compute yourself.
It is trivial to solve your equation for $γ$ as a function 
of the other values.  We get
$$
γ = \sqrt{(\biggr( \frac{C}{v(ω)} \biggl) ^2 - {(ω^2-ω_0^2)^2}) / {ω^2}}
$$

It is almost true that you can substitute in your values and get
several estimates for $γ$, which you can simply average to get a good estimate. 
I say almost true because as $ω$ approaches zero, both the numerator and denominator 
go to zero, so the calculation becomes unstable. As long as you stay away from 
$ω=0$, this should give good results.   @Esmailian provided some nice data in his 
answer. Using his data I get the following estimates of $γ$. 

6.012568  5.999664  5.994524  5.991173  6.009153  5.996748  6.019433
5.998072  6.001754  5.993656  6.002409  5.985938  5.996226  5.998363
6.474674  6.525602 14.140690  5.999124  6.000368  5.998427


Three of his $ω$'s are less than 1. These are the ones that give the estimates of $γ$

6.474674  6.525602 14.140690      


The mean of the remaining estimates is 5.999859. As long as you have enough 
$ω$'s that are not near zero,  this should provide a good estimate of $γ$.