Stacking Cubes

let $S_A$ be the surface area of cuboid $A$ , $S_B$ be the surface area of cuboid $B$ and $x$ be the side length of the small cube, then we have

$S_A=2[x(2x)+2x(2x)+x(2x)=2(2x^2+4x^2+2x^2)=2(8x^2)=16x^2$

$S_B=2[x(4x)+x(x)+x(4x)]=2(4x^2+x^2+4x^2)=2(9x^2)=18x^2$

We know that

$S_B=288+S_A$

So, we have

$18x^2=288+16x^2$ $\large \implies$ $x^2=144$ $\large \implies$ $x=12$

The volume of cuboid $A$ is therefore,

$V_A=x(2x)(2x)=12(24)(24)=$ $\boxed{\color{#D61F06}\large 6912}$

Cuboid A has 16 faces, Cuboid B has 18 faces; since B is 288 units larger than A, 2 faces = 288, 1 face = 144. If the side is x, then x^2=144; and x=12.

Therefore Cuboid A is 12x24x24 = 6912

Sidelength is x then $18x^{2}=16x^2+288$ , solving for x shows that x = 12. The volume is equal to $4x^2=6912$

If the area of an side of a small cube is x, the equation will be, 18x=16x+288 x=144 as x is a square,the side length is 12 so,the volume of small cube is 1728 and the volume of A is 4*1728 =6912