Answer
$\Delta d_{\max }=0.51-0.34=0.17m$
Work Step by Step
$$\Delta P=\rho g\Delta h\Rightarrow \Delta h=\dfrac {\Delta P}{\rho g}=\dfrac {0.05\times 10^{5}}{1.5\times 10^{3}\times 9.8}\dfrac {0.05\times 10^{5}}{1.5\times 10^{3}\times 9.8}\approx 0.34 ( for Dead Sea)$$
$$\Delta P=\rho g\Delta h\Rightarrow \Delta h=\dfrac {\Delta P}{\rho g}=\dfrac {0.05\times 10^{5}}{1\times 10^{3}\times 9.8}\approx 0.51( for Normal Water)$$
So the d max difference is
$$\Delta d_{\max }=0.51-0.34=0.17m$$