Answer
$P_{1gage}=48.90kPa$
Work Step by Step
Starting at point 1:
$P_{1}+\rho_{water}*g*h_{1}+\rho_{oil}*g*h_{2}-\rho_{mercury}*g*h_{3}=P_{2}$
As $P_{2}=P_{atm}$
$P_{1}-P_{atm}=-\rho_{water}*g*h_{1}-\rho_{oil}*g*h_{2}+\rho_{mercury}*g*h_{3}$
As $P_{1}-P_{atm}=P_{1gage}$
$P_{1gage}=g*(-\rho_{water}*h_{1}-\rho_{oil}*h_{2}+\rho_{mercury}*h_{3})$
Substituting:
$P_{1gage}=(9.81\frac{m}{s^2})*[-(1000\frac{kg}{m^3})*(0.2m)-(850\frac{kg}{m^3})*(0.3m)+(13600\frac{kg}{m^3})*(0.4m)]$
$P_{1gage}=48902.85Pa=48.90kPa$
