This problem has two "failure modes", and it is your goal to find the force that would hypothetically cause each. Which ever force is the minimum of the two, that is the maximum allowable force that this member can withstand without failing.
Use the definition of stress, to solve for the force to cause failure by "too much stress":
sigma = Fs/A
Fs = sigma_max*D^2/4
Now elongation to solve for the stress to cause failure by elongation.
Elongation percentage (also known as strain, when not expressed as a percent):
e = (L - L0)/L0
Relationship between stress and strain:
sigma = E*e
Where E is Young's Modulus.
Note: this only holds true when the member is loaded in its linear elastic regime.
Use force of deflection (Fd) to define this stress:
sigma = Fd/A
Fd/A = E*e
Solve for Fd:
Fd = E*e*A
Fd = E*e*pi*D^2/4
Conclusion formulas to calculate:
Fd = E*e*pi*D^2/4
Fs = sigma_max*D^2/4
Data:
sigma_max = 175e6 Pa
e = 0.0014
D = 50e-3 meters
E = 210e9 GPa, this is one you need to look up. Note, for stiffness properties, it usually doesn't matter what variant of carbon steel you use.
Results:
Fs = 343611 Newtons
Fd = 577267 Newtons
We have to select the minimum of the two. Because we want to verify that failure occurs in neither failure mode.
Thus the answer is 343611 Newtons, which translates to 77246 lbs.