If the winner wins on his second draw then 3 balls have been drawn, 2 by player A and one by player B.
A draws. P(RED) on first draw is 3/8, P(WHITE)=5/8. After A has drawn a red, there are 7 balls left, 5W+2R. B draws: P(RED)=2/7, P(WHITE)=5/7. Now A draws again. There are only 6 balls left, and if B has drawn red, then there is only one red left, so P(RED)=1/6; if B drew white, there are still two reds left, so P(RED)=2/6. So the results of drawing resulting in A as the winner are RWR (3/8*5/7*2/6=5/56) or RRR (3/8*2/7*1/6=1/56). Either possibility makes A the winner so P=5/56+1/56=6/56=3/28.