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

$26 H=h$