$2 P_0=P_0+\rho g h$

$\Rightarrow P_0=\rho g h$

$\Rightarrow P_0=75 \,cm \text { mercury } \quad \text { [Atmospheric pressure] }$

$\Rightarrow \rho_{\text {mercury }} \times g \times \frac{75}{100}=\rho_{\text {water }} \times g \times h$

$\Rightarrow \frac{\rho_m}{\rho_w} \times \frac{75}{100}=h$

$\Rightarrow \frac{40}{3} \times \frac{75}{100}=h \quad\left[\because \frac{\rho_m}{\rho_w}=\frac{40}{3} \text { (given) }\right]$

$\Rightarrow h=10 \,m$