Explanation

(b) Pressure at bottom of the lake $= P_{0} + h\rho g$
Pressure at half the depth of a lake $= P_{0} + \frac{h}{2}\rho g$
According to given condition
$P_{0} + \frac{1}{2}h\rho g = \frac{2}{3}\left( P_{0} + h\rho g \right) \Rightarrow \frac{1}{3}P_{0} = \frac{1}{6}h\rho g$
$\Rightarrow h = \frac{2P_{0}}{\rho g} = \frac{2 \times 10^{5}}{10^{3} \times 10} = 20m$