Answer
The required gauge pressure at the heart is $~~1.0\times 10^3~torr$
Work Step by Step
We can find the pressure difference between the head and the heart:
$\Delta P = \rho g \Delta h$
$\Delta P = (1.06\times 10^3~kg/m^3)(9.8~m/s^2)(21~m-9.0~m)$
$\Delta P = 1.25\times 10^5~N/m^2$
$\Delta P = (1.25\times 10^5~N/m^2)(\frac{7.50\times 10^{-3}~torr}{1~Pa})$
$\Delta P = 938~torr$
We can find the required gauge pressure at the heart:
$p_g = 80~torr+938~torr$
$p_g = 1.0\times 10^3~torr$
The required gauge pressure at the heart is $~~1.0\times 10^3~torr$