Thermal expansion would be \(=L \propto \Delta T\)

Where \(L=\) original length

\(\alpha=\) coefficient of linear expansion

\(\Delta T=\) Change in temperature

So substituting values

\(\Delta L=2 \times 10^{-6} \times 80\)

\(\Delta L=1.6 \times 10^{-4} \,m\)

Now \(\Delta L=\frac{F L}{A Y}\)

\(\frac{\Delta L \times A Y}{L}=F\)

Substitute values

\(\frac{1.6 \times 10^{-4} \times 10^{10} \times 1}{2 \times 10000}=F\)

\(80 \,N = F\)