The following graphical solution relies on the idea that it's possible to separate the contribute of the zero in w=0 by referring the magnitude measurements to a line slanted with a -20dB/decade slope. Thanks to the properties of logarithms, division becomes translation on the magnitude Bode plot. We also have to take into account the -90 degrees contribution in the phase - it's basically a constant -90° addition since, being the pole in the origin, it has already 'run its course'.

- the first step is to find wn. In a second order system with no zeros, the phase resonance happens exactly at wn, the undamped natural frequency (a frequency that is in general different from wpeak, the peak frequency of the magnitude, and also from the damped natural frequency wd). Since we need to separate the phase contribution of the pole in the origin, instead of finding the frequency where the phase is -90° we need to find the frequency where the phase is -180°
By eyeballing the scale on the tiny plot I have I believe I can locate it at 6.7 rad/s. You might end up with a better estimate.
Now I want to find the 3dB corner frequency the system would have without the pole in the origin . I therefore...
...trace a slanted line, translated 3dB under the asymptotic behavior at low frequencies and
...and then I look for the intersection of said line with the magnitude of the transfer function to find w3dB. Eyeballing again I find w3dB = 8.7 rad/sec
We are now in the position to computed the ratio w3dB/wn = 1.298 = 1.3
We can now either solve the expression for w3dB as a function of zeta

or, if we have a graph like this, 5) use it to find the value of zeta corresponding to w3dB/wn = 1.3.

again by eyeballing I get a zeta value of around 0.48, a value not dissimilar from that found by solving the equation

for zeta, which gives zeta = 0.477
And this value is reasonably close, considering the amount of lazy eyeballing employed, to the correct answer of 0.447. Try your estimates on a bigger graph, counting the pixels and report back. Did it work?
Caveat emptor: it is imperative that the second order function be without additional zeroes (apart for the one we have been able to separate). The relevant frequencies have different expressions form system with one or more zeroes.