The cross section of the tube of a mercury barometer is 1 square centimeter, and when the barometric height is 760 millimeters, 6 centimeters of the upper end of the tube is empty. When a bubble of air is passed up into the tube it is found that the mercury surface in the tube drops 2 centimeters. Calculate the volume of the bubble of air at atmospheric pressure.

This is what I have:
Hydrostatic Pressure + the Pressure of the air inside the tube = atmospheric pressure
Vo=P/Po
The volume of the air inside the glass tube is V=(6+2) cm^3 and its pressure is P. At the atmospheric pressure, this amount of air would occupy the volume Vo=(P/Po )*V=2/76 *8 cm^3.
13600=Density of Water

Po=0.76 (m) *13600 (kg/m^3) * 9.8 (m/s^2 )= 1.013 *10^5 Pa
P=0.02 (m) * 13600 (kg/m^3)*9.8 (m/s^2) = 2.666*10^3 Pa

P/Po=0.026

But now I'm stuck on where to go from here