There are $6$ cards labeled $1,2,3,4,5,6$.
Two cards are drawn one at a time, so order matters.
There are $6 \times 5 = 30$ ordered ways to draw two cards.
We want the first card to be exactly $2$ greater than the second card.
So if the second card is $1$, the first must be $3$.
If the second card is $2$, the first must be $4$.
If the second card is $3$, the first must be $5$.
If the second card is $4$, the first must be $6$.
The favorable ordered pairs are $(3,1), (4,2), (5,3), (6,4)$
So there are $4$ favorable outcomes.
Thus the probability is $\dfrac{4}{30} = \dfrac{2}{15}.$
Answer: $\boxed{\text{B. } \dfrac{2}{15}}$
