Answer
$P_{gasoline}=164.61kPa$
Work Step by Step
Assuming that the effect of air on pressure is negligible:
$P_{gasoline}+\rho_{gasoline}gh_{gasoline}+\rho_{mercury}gh_{mercury}-\rho_{oil}gh_{oil}+\rho_{water}gh_{water}=P_{gage}$
$P_{gasoline}=P_{gage}-\rho_{gasoline}gh_{gasoline}-\rho_{mercury}gh_{mercury}+\rho_{oil}gh_{oil}-\rho_{water}gh_{water}$
$P_{gasoline}=P_{gage}+g\rho_{water}(-0.70h_{gasoline}-13.6h_{mercury}+0.79h_{oil}-h_{water})$
$h_{gasoline}=22cm*(\frac{1m}{100cm})=0.22m$
$h_{mercury}=10cm*(\frac{1m}{100cm})=0.10m$
$h_{oil}=50cm*(\frac{1m}{100cm})=0.50m$
$h_{water}=45cm*(\frac{1m}{100cm})=0.45m$
Substituting:
$P_{gasoline}=180000Pa+9.81\frac{m}{s^2}*1000\frac{kg}{m^3}(-0.70*0.22m-13.6*0.10m+0.79*0.50m-0.45m)$
$P_{gasoline}=164.61kPa$
