A does half as much work as B in one -sixth of the time.If together they take 10 days to complete a work, how much time shall B take to do it alone?

To do half of the work in one sixth of the time means A is 3 times as efficient as B. To understand this point just think if same work would have been done in one sixth of the time then A would have been six times as efficient as B.
Now they together complete the work in 10 days.
B is working with someone who is 3 times as efficient as himself.
That means 4 people of B's efficiency finished the work in 10 days.
So B alone would have done it in 40 days.
A alone would have taken one third of this time.

Alternatively,
Given,
$${\text{A}} \times \frac{1}{6} = {\text{B}} \times \frac{1}{2}$$
So, A = 3B
Now,
Given they together complete the work in 10 days
So, One Day''s work of,
(A + B) = $$\frac{{100}}{{10}}$$ = 10%
(3B + B) = 10%
4B = 10%
So, one day work of B = $$\frac{{10}}{4}$$ = 2.5%
So, B can complete 100% work in = $$\frac{100}{2.5}$$ = 40 days