Three cubes whose edges measure 3 cm, 4 cm, and 5 cm respectively are melted to form a new cube.

Three cubes of edges $3 \mathrm{~cm}, 4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ are melted and molded to form a new cube.

i. e., volume of the new cube = sum of the volumes of the three cubes

$=(3)^{3}+(4)^{3}+(5)^{3}$

$=27+64+125$

$=216 \mathrm{~cm}^{3}$

We know that volume of a cube $=(\text { side })^{3}$

$\Rightarrow 216=(\text { side })^{3}$

$\Rightarrow$ Side of the new cube $=\sqrt[3]{216}=6 \mathrm{~cm}$

$\therefore$ Surface area of the new cube $=6 \times(\text { side })^{2}=6 \times(6)^{2}=216 \mathrm{~cm}^{2}$