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**Hint**: The given problem is a work and time problem. We will use the unitary method to solve this problem. We are given the time consumed by A to finish a work and the time consumed by A and B together to finish the work. Using these given things we are asked to find the time consumed by B alone to finish the work. We will also introduce algebraic expressions to solve this problem.

**Complete step-by-step answer**:

A can do a piece of work in $ 8 $ days.

Let the work done by A in one day is $ \dfrac{1}{8} $ th of the work done.

A and B together can do the same work in 6 days.

So, the work done by A alone is $ 6 \times \dfrac{1}{8} $ of the total work.

$ 6 \times \dfrac{1}{8} = \dfrac{6}{8} $

Now the rest work be done by B let it be $ x $ .

So, $ \dfrac{6}{8} + x = 1 $

$ \Rightarrow x = 1 - \dfrac{6}{8} = \dfrac{2}{8} $

The work is for $ 6 $ days.

So, the actual work done by B for 1 day is $ \dfrac{1}{6} \times \dfrac{2}{8} = \dfrac{1}{{24}} $

Since, B takes $ 1 $ a day to complete $ \dfrac{1}{{24}} $ part of the work.

This means B takes $ 24 $ days to finish the work alone.

**So, the correct answer is “$ 24 $”.**

**Note**: The given problem is based upon a statement containing some information which we have used to find an algebraic expression. So, formulate the question properly to calculate the result. The result we are obtaining is for $ 6 $ days which we still have to calculate for one day and this can be done by multiplying the reciprocal of $ 6 $ in it.

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