Answer : Option B

Explanation :

ATQ,

A+B = 12 days
B+C = 16 days

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (12 and 16) = 48
Let the total work = 48

Note: One day work = (total work/ days)

Now,

(A+B)'s one day work = 48/12 = 4 unit
(B+C)'s one day work = 48/16 = 3 unit

As per the question:

A works for 5 days
B works for 7 days or (5 + 2) days, that means B works 5 days with A and remaining 2 days with C.
C works 13 days or (2+11) days, that means C works 2 days with B and remaining 11 days alone.

That means total work done by (A+B) in 5 days
A+B = 5 days * 4 unit = 20 units
And, total work done by (B+C) = 2 days * 3 unit = 6 units
So, A+B+C finish the 26 units of work.

Remaining work = 48-26 = 22 unit work
And C completes the remaining work in 11 days.
i.e., C's one day's works = 22/11 = 2 units.

C alone can finish total work in [total work/ C's one day work] = [48/2] = 24 days.