20 log (p_water/p_air) = 20 log (20 µPa/1 µPa) = + 26 dB

Therefore pressure measurements of equal pressures in air and water differ by 26 dB

The characteristic impedance of water is about 3600 times that of air, so the conversion factor for the intensity of sounds of equal pressure in air vs water is 36 dB:

10 log (3600) = 36 dB

If the different reference pressures (1 µPa and 20 µPa) are taken into account, the difference is

36 dB + 26 dB = 62 dB

Therefore intensity measurements of equal pressures in air and water differ by 62 dB. Note that all of these conversions simply relate underwater sounds to those in air. How an animal perceives or reacts to an underwater sound may be very different from its reaction to airborne sounds. For many marine mammals, especially the large cetaceans, there are no established audiograms - in other words, we're not sure of the hearing range of many whales. It is generally assumed, however, that animals can hear the ranges of sounds that they produce.

A simplified example....

If a jet engine has an intensity of 140 dB re 20µPa @ 1m, then underwater the equivalent intensity would be

SIL_water = SIL_air + 62 dB= 202 dB re 1 µPa

To convert from water to air, simply subtract the 62 dB from the SIL in water. A supertanker generating a 190 dB sound level would be roughly equivalent to a 127 dB sound in air. (Note that these are gross generalizations because the source level often changes with the frequency component of the sound.)