### Solution:

We have to find the number of days B will take to finish a work that is twice as big as the earlier one.If a person can do a work in x days, then, we can say that in 1 day that person can do 1/x.

A's 1 day's work =$\frac{1}{24}$

We will now find the number of days required by B to do the same work as A. We know that B is 60% more efficient than A. So, the amount of work done by B in one day can be calculated as follows,

B's 1 day's work =$\frac{1}{24}$+(60% of $\frac{1}{24})$)= $\frac{1}{24}$ ×$\frac{160}{100}$=$\frac{1}{15}$

B can do same work in 15 days

Therefore, the number of days required by B to complete the same quantity of work as A is 15 days.

Now, we are asked to find the number of days B will take to do a work which is twice as large as the earlier one. So, we will multiply the days taken by B to finish the earlier work by 2.

We know that the number of days required by B to finish the original work is 15 days. So, to finish a work which is 2 times the previous work

B can do double work in2×15 = 30 days

Hence, Option (4) is the correct answer.