Answer
$$p_{total} \approx 101.7 \space kPa$$
Work Step by Step
1. Calculate the hydrostatic pressure exerted by the column of water:
** Since the column of water on the open side is lower than the same on the side connected to the apparatus, the height is negative in the calculations.
$$p_{H_2O} = \rho gh = (0.99707 \space g \space cm^{-3})(9.81 \space ms^{-2})(-10.0 \space cm)$$$$p_{H_2O} = -97.8126 \space g \space cm^{-2} \space ms^{-2} \times \frac{1 \space kg}{1000 \space g} \times \Big(\frac{100 \space cm}{1 \space m}\Big)^2$$$$p_{H_2O} = -978.126 \space kg \space m^{-1}s^{-2} \approx -0.978 \space kPa$$
2. Find the total pressure:
$$p_{total} = p_{ext} + p_{H_2O} = 770 \space Torr - 0.978 \space kPa$$
$p_{ext} = 770 \space Torr \times \frac{0.133322 \space kPa}{1 \space Torr} = 102.658 \space kPa$
$$p_{total} = 102.658 \space kPa - 0.978 \space kPa = 101.68 \space kPa$$ $$p_{total} \approx 101.7 \space kPa$$
