First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:

where
v is the wave speed

its wavelength
f its frequency
Using

and

, we get

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:

and since we know both v and t, we can find the distance L between the dolphin and the tuna:
