Use reciprocal rates with careful signs, since a filler adds its rate and a drain subtracts. Let b be pipe B's solo time, so pipe A's is b + 6. The two scenarios give:
Pipe B with the drain: 1/b − 1/d = 1/8.
Pipe A with the drain: 1/(b + 6) − 1/d = 1/24.
Subtract to eliminate the drain: 1/b − 1/(b + 6) = 1/8 − 1/24 = 1/12. The left side is 6 / (b(b + 6)), so b(b + 6) = 72, that is b² + 6b − 72 = 0, factoring as (b − 6)(b + 12) = 0. Only b = 6 is valid, so B fills in 6 hours and A in 12 hours.
Back out the drain: 1/d = 1/6 − 1/8 = 1/24, so the drain empties a full reservoir in 24 hours. Checking pipe A with the drain, 1/12 − 1/24 = 1/24, matches the stated 24 hours.
With all three open, the net rate is 1/6 + 1/12 − 1/24 = 4/24 + 2/24 − 1/24 = 5/24 per hour, so the fill time is 24/5 hours, that is 4.8 hours. So the answer is C.
A (24/7) flips the drain's sign and adds 1/24 instead of subtracting it. B (4) drops the drain and uses only the inlets, though the drain is open in the asked scenario. D (6) reaches the right system but reports pipe B's solo time. E (8) just restates the given pipe-B-with-drain scenario. Each given equation involves the drain rate, so verifying any candidate forces solving the full system.