The maximum elongation that is acceptable for segment $\left(2\right)$ is ${\mathrm{\delta }}_2=2.14 \mathrm{mm}$. Rearrange the axial force-deformation equation to solve for the minimum acceptable area ${\mathrm{A}}_2$.

${\mathrm{\delta }}_2=\frac{{\mathrm{F}}_2{\mathrm{L}}_2}{{\mathrm{A}}_2{\mathrm{E}}_2}$

$\therefore {\mathrm{A}}_2=\frac{{\mathrm{F}}_2{\mathrm{L}}_2}{{\mathrm{\delta }}_2{\mathrm{E}}_2}=\frac{(50 \mathrm{kN})(800 \mathrm{mm})(1,000 \mathrm{N}/\mathrm{kN})}{(2.14 \mathrm{mm})(70,000 \mathrm{N}/{\mathrm{mm}}^2)}=267.02 {\mathrm{mm}}^2$