The water in a tank is pressurized by air, and the pressure is measured by a multi-fluid manometer as shown. determine the gage pressure of air in the tank if h1 = 0.2 m, h2 = 0.3 m, and h3 = 0.46 m. take the densities of water, oil, and mercury to be 1000 kg/m3 , 850 kg/m3 , and 13,600 kg/m3 , respectively

There is something wrong with the given, the second should be 3, 850 kg/m3. So here is the answer for this question:
The formula is P1 + Pwatergh1 + Poilgh2 - Pmercurygh3 = Patm
In order to solve for P1 we need to rearrange the formula, so it will look like:P1 = P atm - Pwatergh1 - Poilgh2 + Pmercurygh3
Noting that P1gage would be equal to P1 - Patm and we can substitute the given above:P1,gage = (9.81 m/s^2)[13,600 kg/m^3)(0.46m) - (1000kg/m^3)(0.2 m)
= (850kg/m^3)(0.3m) (1N / 1kg * m/s^2) (1kPa/ 1000 N/m^2)
= 56.9 kPa