Answer
$4.2\times10^{4}\,N$
Work Step by Step
Given/Known: Original length $L_{0}=3.6\,m$
Change in length $\Delta L=5.7\times10^{-7}\,m$
Area $A= \pi r^{2}=\pi\times(65\times10^{-2}\,m)^{2}$
Young's modulus for steel $Y=2.0\times10^{11}\,N/m^{2}$
Recall: $F=Y(\frac{\Delta L}{L_{0}})A$
Substitute:
$F=2.0\times10^{11}\,N/m^{2}\times\frac{5.7\times10^{-7}\,m}{3.6\,m}\times\pi\times(65\times10^{-2}\,m)^{2}=4.2\times10^{4}\,N$
Force applied on the top of the cylinder is equal to the weight of the object, which is equal to $4.2\times10^{4}\,N$
