Answer (B): Without loss of generality, assume that the first roll is $A^+$. In order for the bee to traverse four distinct edges of a cube, the second roll cannot be $A^+$ or $A^-$, so there are $4$ rolls ($B^+$, $B^-$, $C^+$, and $C^-$) that are allowed at this stage. Each new roll must represent a perpendicular direction for the bee, and there are $3$ choices that remain in compliance for the third roll—$2$ of which extend into three dimensions, and $1$ of which creates a “C” shape. In the former case, there are $2$ choices for the fourth roll, while in the latter case, there are $3$ choices (including the one where the bee traverses four edges forming a square). In total, the number of compliant paths is $6 \cdot 4 \cdot (2 \cdot 2 + 1 \cdot 3) = 2^3 \cdot 3 \cdot 7$. The total number of paths is $6^4 = 2^4 \cdot 3^4$, and the probability that the path represents exactly four edges of a unit cube is
$$\frac{2^3 \cdot 3 \cdot 7}{2^4 \cdot 3^4} = \frac{7}{2 \cdot 3^3} = \frac{7}{54}.$$