Here's my guess at how to look at this problem.
You have a rectangular solid of 16 cubes forming the base. The whole thing is replicated 20 cubes high. At the top, you have 320 GPa of something weighing down on the cubes.
First, you have 320/16 = 20GPa on each vertical pile of cubes. I think you can ignore lateral forces because by the time they get invoked the cube is already crushed.
So the problem is now 20GPa on top of 20 cubes, each of which will crush at a some pressure scattered around 25GPa (your probability distribution). If the cubes didn't weigh anything at all, then all you need to do is find the probability that all 20 cubes crush at >20GPa. That's not too hard - if they're independent identically distributed variables you can calculate the probability for one and then raise that to a power.
My problem with this from a physics perspective is that the cubes must weigh something. That means that the force on the bottom of the column is not only what you have put on top, but also the combined weight of the other 19 cubes. Similarly the next cube has the weight of the top and the other 18 cubes, etc. So you have 20 different probabilities you need to calculate to ensure that all of them don't crush.
Your problem does not mention the weight of a cube anywhere. Is it incomplete?