We will use a formula, but only a very basic one, namely the formula that defines conditional probability. Let $P$ be the event the person tests positive for AIDS, and let $A$ be the event that the person actually has AIDS. We want $\Pr(A|P)$. By the definition of conditional probability, we have
We need to find the two probabilities on the right.
Let us first tackle the harder one, namely $\Pr(P)$. We can test positive in two disjoint ways: (i) We actually have AIDS, and test positive or (ii) We don't have AIDS, yet test positive.
For (i), the probability we have AIDS is $0.003$. Given we have AIDS, the probability of testing positive is $0.995$. So the probability of (i) is $(0.003)(0.995)$.
For (ii), the probability we don't have AIDS is $0.997$. Given we don't have AIDS, the probability we test positive is $0.02$. Note: Here we are going against the wording of the problem. It says that if we don't have AIDS, the probability we still test positive is $0.98$. This is such a ridiculously implausible figure that surely what is intended is that the probability we test
negative is $0.98$.
So, with this reinterpretation of the figures, the probability of (ii) is $(0.997)(0.02)$. Thus
We can think of the above expression for $\Pr(P)$ as coming from the law of total probability. However, the formula that you quote is not correct for this problem.
There is a general formula. I would prefer not to give it, since figuring out things each time from basics is better for you. You are much more likely to know what's going on, and, very importantly, you are much more likely to get the right answer.
OK, enough preaching. We still need $\Pr(A\cap P)$. But that's already done, it is the probability of (i). We conclude that
Again, please remember that we have reinterpreted the given figures as saying that the AIDS test has probability $0.02$ of giving a false positive. The actual wording of the problem, surely wrong, seems to say that the probability of a false positive is $98\%$, a ridiculously implausible number.