Solution:

Let $X$, $Y$, $Z$ be the three directions in which the moth can initially go. We can symbolize the trajectory of the moth by a sequence of $X$'s, $Y$'s, and $Z$'s in the obvious way: whenever the moth takes a step in a direction parallel or opposite to $X$, we write down $X$, and so on.

The moth can reach $B$ in either exactly $3$ or exactly $5$ steps. A path of length $3$ must be symbolized by $X Y Z$ in some order. There are $3! = 6$ such orders.

A trajectory of length $5$ must be symbolized by $X Y Z X X$, $X Y Z Y Y$, or $X Y Z Z Z$, in some order. There are $3 \cdot \frac{5!}{3!1!1!} = 3 \cdot 20 = 60$ possibilities here. However, we must remember to subtract out those trajectories that already arrive at $B$ by the $3$rd step: there are $3 \cdot 6 = 18$ of those.

The answer is thus $60 - 18 + 6 = 48$.