Each of the $12$ edges of a cube is labeled $0$ or $1$. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$?
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$
For simplicity, we will name this cube $ABCDEFGH$ by vertices, as shown below.

Note that for each face of this cube, two edges are labeled $0,$ and two edges are labeled $1.$ For all twelve edges of this cube, we conclude that six edges are labeled $0,$ and six edges are labeled $1.$
We apply casework to face $ABCD.$ Recall that there are $\binom42=6$ ways to label its edges:


Opposite edges have the same label.
There are $2$ ways to label the edges of $ABCD.$ We will consider one of the ways, then multiply the count by $2.$ Without loss of generality, we assume that $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ are labeled $1,0,1,0,$ respectively:
We apply casework to the label of $\overline{AE},$ as shown below.

We have $2\cdot2=4$ such labelings for this case.
  Opposite edges have different labels.
There are $4$ ways to label the edges of $ABCD.$ We will consider one of the ways, then multiply the count by $4.$ Without loss of generality, we assume that $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ are labeled $1,1,0,0,$ respectively:
We apply casework to the labels of $\overline{AE}$ and $\overline{BF},$ as shown below.

We have $4\cdot4=16$ such labelings for this case.

Therefore, we have $4+16=\boxed{\textbf{(E) } 20}$ such labelings in total.
~MRENTHUSIASM