Fres,
Something is wrong with your numbers. Assume the 24VDC is rms. Then you are getting
20 * log(50e-9/24) = -173.62 dB attenuation.That is not going to happen on a pcb. It requires metallic enclosure with extremely low leakage and no conductors penetrating the enclosure.
Even good 50 ohm coax has leakage on the order of -90 dB. So you are making measurements that are 173-90 = 83 dB better than coax leakage. That does not seem possible.
Now assume the 50nV is rms. This amount of rms noise in a 400KHz bandwidth requires a 0.379 ohm resistor at 25 degrees C. Your input impedance is much higher than that, and you are unlikely to be using transistors with zero noise.
The 98% efficiency number seems high.
Mike