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

(b) $144 \mathrm{~cm}^{2}$

Volume of the new cube formed = total volume of the three cubes
Suppose that a cm is the edge of the new cube, then

$a^{3}=3^{3}+4^{3}+5^{3}$

$=27+64+125$

$=216$

$=6 \mathrm{~cm}$

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