When A goes first, he has a probability of 1/3 finding $100 on the first try. If he fails (p=2/3), then B has a probability of 1/2 of finding the $100 his try, so overall probability for B is 2/3 x 1/2 = 1/3. And if B fails (p = 2/3 x 1/2 = 1/3) then A opens last box and finds the $100. So overall, when A goes first he finds the $100 with p = 1/3 + 1/3 = 2/3.
The opposite is the case when B goes first.
After the first three games, the expected winnings for A = 2/3 x $300= $200 and expected winnings for B = $100.
The opposite is the case for games 4, 5, and 6, so after game 6 the expected winnings for A will be $300 and the same for B.
In game 7, B goes first again and has p = 2/3 chance of winning the $100.
So overall, B has the best chance of winning $400, and thus gets the first chance of winning the car.