A vertical off-shore structure is built to withstand a maximum stress of ${10^9}Pa$ . Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly $3k.m.$ , and ignore ocean current?

Note: To solve this problem one should know about the basics of fluid statics. Here fluid is let to stable and the object is dipped in the ocean so water of the ocean exerted pressure on the object.
Fluid pressure: fluid pressure within a point submerged in fluid. The pressure arises due to the weight of the fluid above the point. As fluid doesn't have any shape it can change its form according to the container. Unit of pressure is Pascal $Pa$ .
Formula used:
$p = {p_a} + \rho gh$
Here,
 $p = $ Pressure at a point within the fluid.
 ${p_a} = $ Atmospheric pressure above fluid surface.
 $\rho = $ Density of fluid.
 $g = $ Gravity due to earth.
 $h = $ Distance between fluid surface and point within fluid.

Complete step by step solution:
Given in question,
 $h = 3k.m. = 3000m$
As we know that,
Density of water is $1000kg.{m^{ - 3}}$
 ${p_a} = 1.01 \times {10^5}Pa$
 Substitute all this value in the above maintained equation. We get,
 $p = 1.01 \times {10^5} + 1000 \times 9.8 \times 3000$
 $ \Rightarrow p = 29501000Pa = 0.29 \times {10^9}Pa$
 So, we had found that the pressure exerted on the vertical off-shore structure will be $0.29 \times {10^9}Pa$ .
The maximum pressure up to which it can withstand is ${10^9}Pa$ .
So, we find that the exerted pressure is less than the maximum withstand pressure off vertical off-shore structure. Hence, it is suitable for putting up on top of an oil well in the ocean.

Additional information:
Fluid pressure only depend on:
Density: the pressure on point within the fluid will always depend on the density of the fluid.
Distance of point from the surface of fluid: The pressure only depends upon the depth of point not on volume of water above the point. So, by varying the shape of the container but keeping depth the same we find the pressure at bottom will be the same.
Note: In calculating the fluid pressure we assume that it is point size structure. We can calculate the density of fluid by using volume and weight of given fluid.
 $\rho = \dfrac{W}{V}$