This answer requires some spatial examples, so I think it will be better if I generalize the case of a simple cube, which can be applied to all possible situations.
As you said, the total area covered by the paint in a cube with lenght $a$ is equal to:
$$
S_{\mathrm{ext}}=6\cdot a^2
$$
But then the division into $27$ cubes with lenght $1/3$ of the original gives the new surface area, greater than the original because new slices prodce more internal surface; so now the total surface is expressed by:
$$
S_{\mathrm{tot}}=27\cdot 6\cdot (a/3)^2=\frac{27\cdot 6}{9}\cdot a^2=3\cdot 6\cdot a^2=3S_{\mathrm{ext}}
$$
Now think about a Rubik's Cube, with only the cubes facing the outside are painted: since you know that the total amount of paint occupies the external surface $S_{\mathrm{ext}}$, then to know the surface not covered in paint you have to simply subtract to the total sliced surface $S_{\mathrm{tot}}$ the external one:
$$
S_{\mathrm{unpaint}}=S_{\mathrm{tot}}-S_{\mathrm{ext}}=3S_{\mathrm{ext}}-S_{\mathrm{ext}}=2S_{\mathrm{ext}}=12\cdot a^2
$$
Now you have basically derived the sum of the internal surface of every little cube, because the sum of the external is nothing more than the total covered in paint, or $S_{\mathrm{ext}}$ (with the Rubik's Cube case, the only cubes coloured).
EDIT: If the number of slices is not defined, it is generalized to a single parameter $n\in \mathbb{N}$, then the new surface tends to be greater as $n$ gets larger.
Let's assume that one cut is repeated on every edge of the cube, so if an edge is divided in three (with two slices), then the number of cubes created is expressed as follows:
$$
N_{\mathrm{cubes}}=(n+1)^3
$$
So if I "slice" a cube for every edge once ($n=1$), I get $N_{\mathrm{cubes}}=(2)^3=8$ little cubes.
The surface of each sub-cube is expressed by:
$$
S_{\mathrm{sub-cb}}=6\cdot\left(\frac{a}{n+1}\right)^2
$$
Then the total surface is expressed as th single one multiplied by the total number of sub-cubes created $N_{\mathrm{cubes}}$:
$$
S_{\mathrm{tot}}=(n+1)^3\cdot 6\cdot\left(\frac{a}{n+1}\right)^2=(n+1)\cdot 6\cdot a^2=(n+1)\cdot S_{\mathrm{ext}}
$$
Now, applying the same deduction as before, the surface not covered with paint is given by a simple subtraction between the total surface and the ponly painted one, which is the external one:
$$
S_{\mathrm{unpaint}}=S_{\mathrm{tot}}-S_{\mathrm{ext}}=(n+1)\cdot S_{\mathrm{ext}}-S_{\mathrm{ext}}=n\cdot S_{\mathrm{ext}}
$$
Now remember that the parameter $n$ is the number of slices, and not the sub-cubes created; as your previous example, the slices were $2$, and so this number appeared inside the final result.
For the third method you need to do faces (1 side painted) - edges (2 sides painted) + vertices (3 sides painted). The idea is that the edges get counted twice by each face so we fix by subtracting that total back out. But then the vertices (corners of the cube) end up not being counted at all since each vertex is hit by 3 faces and 3 edges have we have just done 3 - 3. So we have to add back in the number of vertices.
There are 6 faces with 10*10 cubes which gives 600.
There are 12 edges with 10 cubes which gives 120.
There are 8 vertices (corners) with a single cube each which gives 8.
So we have $600-120+8 = 488$.
Best Answer
Here's my hint: Stare at that for a sufficiently long time, and the solution will come up.
and if that doesn't help, try this: