We know from the isoperimetric inequality that locally the surface must be a sphere (where we can include the plane as the limiting case of a sphere with infinite radius). Also, the surface must be orthogonal to the cube where they meet; if they're not, you can deform the surface locally to reduce its area. A sphere orthogonal to a cube face must have its centre on that face. You can easily show that it can't contain half the volume if it intersects only one or two faces. Thus, it must either intersect at least three adjacent faces, in which case its centre has to be at the vertex where they meet, or it has to intersect at least two opposite faces, in which case it has to be a plane.
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.
Best Answer
I agree with you. An exact calculation is $6(36.25^2-36^2)=108.375\ in^2$ The answer key is close if you are supposed to report the answer in ft$^2$, because $\frac {108}{144}=\frac 34$, but that is still not $\frac 9{16}$