Answer
$P=106.984 KPa$
Work Step by Step
We can determine the required pressure as follows:
$P=P_{\circ}+\Delta P_{oil}+\Delta P_{water}$
$\implies P=P_{\circ}+\rho_{oil} gh_{oil}+\rho_{water}gh_{water}$
We plug in the known values to obtain:
$P=101.3\times 10^3+(0.75\times 10^3\times 9.8\times 4\times 10^{-2})+(10^3\times 9.8\times 55\times 10^{-2})$
This simplifies to:
$P=106.984 KPa$
