Answer : Option B

Explanation :

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

Take the LCM of days = LCM of (10, 15, and 30) = 30
Let the total work = 30

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

Now,

A's one day work = 30/10 = 3 unit
B's one day work = 30/15 = 2 unit
C's one day work = 30/30= 1 unit
Let A works for days = D

Note: man * days = total work

ATQ, (A+B+C) works 6 units work per day till D days, so their total work = 6 * D
(B+C) works 3 unit work per day till 1 day, so their total work = 3 * 1 = 3
C works 1 unit work per day till 3 days, so his total work = 1 * 3 = 3

Or, 6*D + 3*1 + 1*3 = 30 (total work)
6D = 30-6
D = 24/6
D= 4

So, A works for 4 days, B works for 4+1=5 days, and C works for 4+1+3= 8 days.