#### Answer

(a) The gauge pressure at a depth of 250 meters is $2.52\times 10^6~N/m^2$.
(b) The net force on the window is $1.78\times 10^5~N$.

#### Work Step by Step

(a) We can find the gauge pressure at a depth of 250 meters.
$P_G = \rho~g~h$
$P_G = (1030~kg/m^3)(9.80~m/s^2)(250~m)$
$P_G = 2.52\times 10^6~N/m^2$
The gauge pressure at a depth of 250 meters is $2.52\times 10^6~N/m^2$.
(b) The difference in pressure on the outside and inside of the window is the gauge pressure. We can find the net force on the window.
$F = P_G~A$
$F = (2.52\times 10^6~N/m^2)(\pi)(0.150~m)^2$
$F = 1.78\times 10^5~N$
The net force on the window is $1.78\times 10^5~N$.