Cube-Sphere

Let $x$ be the side length of one cube, $S_A$ be the total surface area of the $88$ small cubes, and $S_B$ be the surface area of the cube-sphere.

Then,

$S_A=88(6x^2)=528x^2$

$S_B=6(4x)^2+x(2x)(4)(6)=96x^2+48x^2=144x^2$

Given in the problem that,

$S_A=S_B+384$ $\implies$ $528x^2=144x^2+384$ $\implies$ $384x^2=384$ $\implies$ $x^2=1$

Therefore, $S_B=144x^2=144(1)=$ $\color{#D61F06}\large \boxed{144}$

Let a denote the edge length of the cube. We can see that the TSA of cube sphere is $6\times(24a^2)=144a^2$ while the TSA of 88 cubes is $88\times 6a^2=528a^2$ .
Decrement in SA $=384=(528-144)a^2=384a^2$ $\Rightarrow a=1$ Hence TSA of cube sphere= $144a^2=\Large\boxed{\color{#D61F06}{144}}$ .

The wording of the problem caused some confusion. "When 88 cubes are rearranged " -- rearranged from what?

I supposed that the "rearrangement" could be from 88 single cubes, because it probably wasn't from a 4x4x5 rectangular solid, which is neither a cube nor a sphere.

My work consisted of multiplying the number of cubes (88) times the number of sides of a cube (6), and then subtracting the reduction in surface area (384).

$88 \times 6 - 384 = 144$

A visual inspection of the illustration shows that each "side" of the cube-sphere has an area of 24, times 6, equals 144.

$24 \times 6 = 144$

Let $x$ be the side length of the smaller cubes. Then the total surface area of the $88$ smaller cubes is $6(88)(x^2)=528x^2$ .

The total surface area of the cube-sphere solid is $6(4x)^2+24(x)(2x)=144x^2$

Then we have the equation

$144x^2=528x^2-384$

$x^2=1$

So the total surface area of the cube-sphere is $144x^2=144(1)=\boxed{144}$

Let $x$ be the side length of one small cube. Then our equation is

$88(6x^2)-384=6(4x)^2+4(x)(2x)(6)$

$528x^2-384=96x^2+48x^2$

$384x^2=384$

$x=1$

So the desired surface area is $96+48=\boxed{144}$ .

We let $a$ be the side length of each cube. The surface area of one cube is $6$ $a^2$ . The total surface area of $88$ $cubes$ is $(88)(6a^2)=528a^2$ .

Based from the figure, the total surfaced area of the cube-sphere is $6(4a)^2-6(2a)(2a)+[2(2a)(a)+2(2a)(a)+2a(2a)](6) = 96a^2-24a^2+72a^2=144a^2$ . From the problem, $528a^2-144a^2=384$ , it follows that $a^2=1$ . Therefore, the total surfaced area of the cube-sphere is $144a^2=144(1)=144$