It is true that without knowing the FFT length and sampling frequency, you can't calculate the actual noise level. However, it is easy enough to make an estimate from the just looking at the FFT plot.
Assuming a lenient FFT frequency bin width of 6 Hz (Δf = fs/n_samples, assuming fs = 384 kHz, that would mean at sample length of 64 k samples):
- Eyeballed average grass height = -115 dB
- If reference voltage is 1 voltage unit, noise voltage magnitude of each bin = 10^(-115/20) = 1.78e-6 units
- From 20 - 20000 Hz, there are (20000 -20)/ 6 = 3330 bins
- Equivalent sum of the noise voltage is calculated using root sum of squares: V_sum = sqrt( 3330 * 1.78e-6^2 ) = 1.0e-4 units
- Signal-to-noise ratio = 20 log10(1.0e-4/1.0) = -80 dB (i.e. 35 dB of FFT processing gain)
The signal to noise ratio, with the above assumptions, is just 13 bits! It will be worse if the measurement FFT sample length was longer, which is actually likely to be the case.