#### Answer

The pressure 11 km below sea level is $1.10\times 10^3~atm$

#### Work Step by Step

$P = P_0 + \rho~g~h$
$P$ is the pressure
$P_0$ is the atmospheric pressure
$\rho$ is the density of the liquid
$h$ is the depth below the surface
We can find the pressure 11 km below sea level.
$P = P_0 + \rho~g~h$
$P = (1.01\times 10^5~Pa) + (1030~kg/m^3)(9.80~m/s^2)(11,000~m)$
$P = 1.11\times 10^8~Pa$
We can convert this pressure to units of atmospheres.
$P = (1.11\times 10^8~Pa)(\frac{1~atm}{1.01\times 10^5~Pa})$
$P = 1.10\times 10^3~atm$
The pressure 11 km below sea level is $1.10\times 10^3~atm$.