Answer
$10\,cm^{2}$
Work Step by Step
Surface area of a cube with side of length L is $6\times L^{2}$.
Increase in surface area= $6(L+\Delta L)^{2}-6L^{2}$
$=6(L^{2}+(\Delta L)^{2}+2\times L\times\Delta L)-6L^{2}$
$=6(\Delta L)^{2}+12\times L\times\Delta L$
But $\Delta L=L\alpha \Delta T$
Therefore, the increase in surface area
$=6(L\alpha \Delta T)^{2}+12\times L\times L\alpha \Delta T$
Given/known: Edge length $L=30\,cm$,
Change in temperature $\Delta T=(75-20)^{\circ}C=55^{\circ}C$
Coefficient of linear expansion of copper $\alpha=17\times10^{-6}/C^{\circ}$
Result: Increase in the surface area
$= 6(30\,cm\times17\times10^{-6}/C^{\circ}\times55^{\circ}C)^{2}+12\times30\,cm(30\,cm\times17\times10^{-6}/C^{\circ}\times55^{\circ}C)=10\,cm^{2}$
