A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?

Volume of the large cube = $33+43+53$ = 216 cu.cm. 

Let the edge of the large cube be a. 

So, $a3$= 216 

=>a = 6 cm. 

Required ratio = $\frac{632+42+52}{662}=\frac{25}{18}$