College Physics (4th Edition)

The depth of the lake is $403~m$
We can use $v = 1493~m/s$ as the speed of sound in freshwater at a temperature of $25^{\circ}C$. We can find the total distance the sound travels in the water: $d = v~t = (1493~m/s)(0.540~s) = 806~m$ Since the sound reached the bottom of the lake and returned to the boat, the depth of the water is $\frac{806~m}{2} = 403~m$