Answer
$h=849.47$
Work Step by Step
$750mbars*\frac{0.1kPa}{1mbar}=75kPa$
$650mbars*\frac{0.1kPa}{1mbar}=65kPa$
Assuming $1$ as the beginning and $2$ as the end, and knowing that:
$P=\rho*g*h$
$h_{1}=\frac{P_{1}}{\rho*g}=\frac{75kPa}{1.20\frac{kg}{m^3}*9.81\frac{m}{s^2}}=6371.05m$
$h_{2}=\frac{P_{2}}{\rho*g}=\frac{65kPa}{1.20\frac{kg}{m^3}*9.81\frac{m}{s^2}}=5521.58m$
Then the vertical distance climbed is:
$h_{1}-h_{2}=6371.05m-5521.58m=849.47$
