Please help me to solve this question. Thanks.?

Let y days be the time taken that B finishes the work alone.
Then, time taken that A finishes the work alone = 2y days
and time taken that C finishes the work alone = (2y/3) days

Fraction of work finished in one day :
(1/y) + (1/2y) + (3/2y) = 1/2
(2/2y) + (1/2y) + (3/2y) = 1/2
6/2y = 1/2
2y = 12
y = 6

Time taken that B finishes the work alone = 6 days

A takes twice as much time as B or thrice as much time as C to finish a piece of work.
Working together, they can finish the same work in 2 days.
In how many days can B do the same work alone?
Suppose A, B and C take x, x/2 and x/3 days respectively to finish the work.
Then, 1/x + 2/x + 3/x = 1/2
==> 6/x = 1/2
==> x = 12.
So, B takes (12/2) = 6 days to finish the work.

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