Answer
$1.2\times10^{4}N$
Work Step by Step
Here we use equation 10.17 $F=Y(\frac{\Delta L}{L_{0}})A$ to find the tension of the wire. Where $\Delta L$ is twice the circumference of the peg & considering the wire has a circular cross-section, $A=\pi r_{w}^{2}$.
$F=Y(\frac{4\pi r_{p}}{L_{0}})\pi r_{w}^{2}=Y(\frac{4 r_{p}}{L_{0}})(\pi r_{w})^{2}$ ; Let's plug known values into this equation.
$F=\frac{4(2\times10^{11}N/m^{2})(1.8\times10^{-3}m)}{0.76\space m}[\pi(0.8\times10^{-3}m)]^{2}\approx1.2\times10^{4}N$
