*I don't know how to do this one for sure*, but according to wikipedia, the difference in pressure gives rise to a change in height in a manometer related by

$\displaystyle \Delta h = \frac{\Delta P}{\rho g}$

$\displaystyle \Delta h = \frac{P_{atm} - P_{pipe}}{\rho g}$

$\displaystyle P_{pipe}$ is what you're looking for, but you might need to take into account the difference in area in the left-hand tube by equating force:

$\displaystyle F = PA$

$\displaystyle P_1 A_1 = P_2 A_2$

So you'd get something like

$\displaystyle \Delta h = \frac{P_{atm} - P_2}{\rho g}$

where

$\displaystyle P_2 = \frac{P_{pipe} A_{pipe}}{A_{2}} = 100 P_{pipe}$

So maybe try something like that? Have a go and see what you get

EDIT: Okay, so the above is probably wrong... I'll keep investigating.