A cube is taken and all its six faces are painted. Then, the cube is cut by six planes so as to yield the maximum number of smaller cubes. If 3.5 litres of paint was consumed to paint the original cube, find the additional amount of paint needed to paint all the unpainted faces of all the smaller cubes.

To obtain the maximum number of smaller cubes from a cube by cutting with 6 planes, the planes must be distributed equally parallel to each pair of faces. With 6 planes, this means 2 cuts parallel to the xy-plane, 2 cuts parallel to the yz-plane, and 2 cuts parallel to the xz-plane. This divides the original cube into (2+1) x (2+1) x (2+1) = 3 x 3 x 3 = 27 smaller cubes. Let the side length of the original cube be L. Its surface area is $6L^2$. This area was painted with 3.5 litres. The side length of each smaller cube is L/3. Each smaller cube has a surface area of $6 \times (L/3)^2 = 6L^2/9 = 2L^2/3$. The total surface area of all 27 smaller cubes is $27 \times (2L^2/3) = 18L^2$. The original painted area was $6L^2$. The unpainted area consists of the faces exposed by the cuts. The total surface area of the smaller cubes ($18L^2$) minus the original painted area ($6L^2$) gives the total unpainted area ($12L^2$). The amount of paint is proportional to the surface area. Paint for $6L^2$ litres. Paint for the additional unpainted $12L^2$