#### Answer

The pressure at the bottom of the water layer is $1.17\times 10^5~N/m^2$

#### Work Step by Step

$P = P_0 + \rho~g~h$
$P$ is the pressure
$P_0$ is the atmospheric pressure
$\rho$ is the density of the liquid
$h$ is the depth below the surface
We can find the pressure $P_1$ at the bottom of the oil layer.
$P_1 = P_0 + \rho_o~g~h_1$
$P_1 = (1.013\times 10^5~N/m^2) + (900~kg/m^3)(9.80~m/s^2)(0.50~m)$
$P_1 = 1.057\times 10^5~N/m^2$
We can find the pressure $P_2$ at the bottom of the water layer.
$P_2 = P_1 + \rho_w~g~h_2$
$P_2 = (1.057\times 10^5~N/m^2) + (1000~kg/m^3)(9.80~m/s^2)(1.20~m)$
$P_2 = 1.17\times 10^5~N/m^2$
The pressure at the bottom of the water layer is $1.17\times 10^5~N/m^2$.