Player A goes first, so the chances are not symmetrical.
If player A has 100% accuracy, and goes first, then of course player B has a 0% chance of winning regardless of accuracy.
If player A has 0% accuracy, then player B has 100% chance of winning so long as he has greater than 0% accuracy.*
Now in this case player A has 30% accuracy. You can see that if player B has 100% accuracy, he has a 70% chance of winning. And if player B has 0% accuracy, he has a 0% chance of winning. By varying B's accuracy, you can attain any chance of winning for B between 0% and 70% inclusive.
So please note that it's only possible to solve this question because player A's accuracy is less than (or equal to) 50%. If player A had 51% accuracy, there would be no possible way that player B could have an equal chance of winning.
*I'm going to use "he" as these are hypothetical people and it makes the writing flow better for me.