Let initial radius be \(=r\)

Final radius \(=r+200 \%\) of \(r\)

\(=3 r\)

Atmospheric pressure \(=\rho g H\)

Let depth of the lake be \(h\)

So, pressure at the bottom of lake \(=\rho g H+\rho g h\)

Using \(P_1 V_1=P_2 V_2\)

\(\rho g H \times \frac{4}{3} \pi(3 r)^3=(\rho g H+\rho g h) \times \frac{4}{3} \pi r^3\)

\(\rho g H \times \frac{4}{3} \pi \times 27 r^3=(\rho g H) \frac{4}{3} \pi r^3+\rho g h \times \frac{4}{3} \pi r^3\)

Solving this equation we get

\(26 H=h\)