If three equal cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to the sum of the surface areas of the three cubes will be ?

Let the length of each edge of each cube be a
Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a
Surface area of the cuboid :
$$\eqalign{ & = 2\left[ {3a \times a + a \times a + 3a \times a} \right] \cr & = 2\left[ {3{a^2} + {a^2} + 3{a^2}} \right] \cr & = 14{a^2} \cr} $$
Sum of surface area of 3 cubes :
$$\eqalign{ & = \left( {3 \times 6{a^2}} \right) \cr & = 18{a^2} \cr} $$
∴ Required ratio :
$$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$