When a large bubble rises from the bottom of a lake to the surface, its radius doubles. The atmospheric  pressure is equal to that of a column of water of height H. The depth of the lake is

Let r be the radius of the bubble at the bottom of lake of depth x. The pressure on the bubble is ${\mathrm{\rho }}_{\mathrm{w}}\mathrm{g}(\mathrm{H}+\mathrm{x})$. 

When the bubble rises to the surface of the lake, its radius is 2r and pressure is ${\mathrm{\rho }}_{\mathrm{w}}\mathrm{gH}$. 

Applying Boyle's law, we get
$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{w}}\mathrm{gH}\times \frac{4}{3}\mathrm{\pi }(2\mathrm{r}{)}^3={\mathrm{\rho }}_{\mathrm{w}}\mathrm{g}(\mathrm{H}+\mathrm{x})\times \frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{Solving} \mathrm{we} \mathrm{get}, \mathrm{x} =7\mathrm{H} \end{gathered}$$