Length of the wire, l = 4m
Radius of the wire, r = 4 mm = 0.004 m
Force on the wire, F = 80$\mathrm{\pi } \mathrm{N}$ 
Strain = 0.01% = 0.0001

Now, using the formula of stress,

$\mathrm{Stress} = \frac{\mathrm{F}}{{\mathrm{\pi r}}^2}$  = $\frac{80\mathrm{\pi }}{\mathrm{\pi }(0.004{)}^2} = 5\times {10}^6 \mathrm{N}/{\mathrm{m}}^2$

Young's modulus of the wire is given by, 
Y = $\frac{\mathrm{Stress}}{\mathrm{Strain}}=\frac{5\times {10}^6}{0.0001} = 5\times {10}^{10} N/m^2$ 

Given,

Length of one end of wire, L = 1m

Angular velocity,=12 rad/s

Mass of the load, M=2kg

Therefore tension in the string is,

 
Increase in the length of the wire = (5.53 - 1) = 4.53 m

Given, 
Weight of the load = 200 N
Elongation caused in the wire is given by, $∆l\mathit{ }\mathit{=}\mathit{ }0.8 \mathrm{mm} = 8\times {10}^{-4} \mathrm{m}$ 

To find - Elastic potential energy stored in the wire = ?

Therefore, energy stored in the wire is given by,

 $$\begin{gathered} \mathrm{U}=\frac{1}{2}\mathrm{F}\times \Delta l \\ = \frac{1}{2}\times 200\times 8\times {10}^{-4} \\ = 0.08 J \end{gathered}$$  

The situation is shown in figure.


From right angled 

The different forces acting on mass are:

(i) Weight mg in vertically downward direction. (ii) Tension T in the string along AS. Resolve the components of T as shown in figure. The component 7cosθ balances the weight mg of the mass and component 7sinθ provides the necessary centripetal force to whirl the mass.