Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube.

Three cubes of metal with edges 3cm, 4cm and 5 cm are melted to form a single cube.

∴ Volume of the new cube = sum of the volumes the old cubes

$=\left(3^{3}+4^{3}+5^{3}\right) \mathrm{cm}^{3}$

$=(27+64+125) \mathrm{cm}^{3}=216 \mathrm{~cm}^{3}$

Suppose the edge of the new cube = x cm
Then we have:

Then $216=x^{3}$

$\Rightarrow x=\sqrt[3]{216}=6$

i.e., the edge of the new cube is $6 \mathrm{~cm}$.

$\therefore$ Lateral surface area of the new cube $=4 x^{2} \mathrm{~cm}^{2}=4 \times 6^{2} \mathrm{~cm}^{2}=144 \mathrm{~cm}^{2}$