Answer
75 columns are required.
Work Step by Step
The ultimate strength of steel is $400\times 10^6~N/m^2$
We can find the weight that each column can support such that the stress is $200\times 10^6~N/m^2$:
$W = (200\times 10^6~N/m^2)(0.0960~m^2) = 1.92\times 10^7~N$
In part (a), we found that the total weight of the ground material is $~~1.43\times 10^9~N$
We can find the required number of columns:
$\frac{1.43\times 10^9~N}{1.92\times 10^7~N} = 74.5$
75 columns are required.
