Answer
$233N$
Work Step by Step
We know that
$\Delta F=Y(\frac{\Delta L}{L_{\circ}})A$
We plug in the known values to obtain:
$\Delta F=(2.4\times 10^{10}N/m^2))(\frac{\pi(3.5\times 10^{-3}m)}{0.82m})(\frac{\pi(0.93\times 10^{-3}m)^2}{4})$
$\Delta F=218.7N$
Now the final tension is given as
$F^{\prime}=F+\Delta F$
We plug in the known values to obtain:
$F^{\prime}=14+218.7=233N$
