Answer
$\Delta P=6.4\space KPa$
Work Step by Step
We can determine the required pressure as follows:
$\Delta P=\rho_w gh$
We plug in the known values to obtain:
$\Delta P=10^3\times 9.8\times 65\times 10^{-2}$
This simplifies to:
$\Delta P=6.4\times 10^3Pa$
$\implies \Delta P=6.4\space KPa$
