A and B together can complete a task in 12 days. A alone can complete the task in 18 days. In how many days can B alone complete it?

Let the total amount of work that needs to be done be $$W$$, and the efficiencies of $$A$$ and $$B$$ be $$a$$ and $$b$$, respectively. 

The amount of work done by $$A$$ and $$B$$ together in 12 days is,

$$12(a+b)=W$$ --(1)

The amount of work done by $$A$$ alone in 18 days is,

$$18(a)=W$$

$$a=\dfrac{W}{18}$$ --(2)

Substituting (2) in (1), we get,

$$12\left(\dfrac{W}{18}\ +\ b\right)\ =\ W$$

$$\dfrac{2W}{3}\ +\ 12b\ =\ W$$

$$12b\ =\ \dfrac{W}{3}$$

$$W\ =\ 36b$$

So, $$B$$ alone can complete the total work in 36 days.

Hence, the correct answer is option B.