#### Answer

The submarine's maximum safe depth in seawater is 3.16 km

#### Work Step by Step

We can find the area of the window.
$A = \pi~r^2$
$A = \pi~(0.10~m)^2$
$A = 0.0314~m^2$
We can find the net pressure the window can withstand.
$P_{net} = \frac{F}{A}$
$P_{net} = \frac{1.0\times 10^6~N}{0.0314~m^2}$
$P_{net} = 3.185\times 10^7~N/m^2$
We can find the maximum safe pressure $P$ outside the window if the pressure inside $P_i$ is 1 atm.
$P_{net} = P-P_i$
$P = P_{net}+P_i$
$P = 3.185\times 10^7~N/m^2+1.013\times 10^5~N/m^2$
$P = 3.195\times 10^7~N/m^2$
We can find the depth $h$ below sea level which has this pressure.
$P = P_0 + \rho~g~h$
$h = \frac{P-P_0}{\rho~g}$
$h = \frac{(3.195\times 10^7~N/m^2)-(1.013\times 10^5~N/m^2)}{(1030~kg/m^3)(9.80~m/s^2)}$
$h = 3160~m = 3.16~km$
The submarine's maximum safe depth in seawater is 3.16 km