Let $a$ be the probability that the first player (ultimately) wins if the two players are tied in wins. Let $b$ be the probability that she wins if she is $1$ ahead. And let $c$ be the probability she wins if she is $1$ behind. We have the equations
$$\begin{align}a&=rb+(1-r)c,\\
b&=r+(1-r)a, \\
c&=ra.\end{align}$$
Solve the system of linear equations.
Let $a$ be the expected number of additional games that need to be played if the two players are tied in wins. In particular, $a$ is our required number, since the players start off tied.
Let $b$ be the expected number of additional games if Player $1$ is leading by $1$. And let $c$ be expected number of additional games if Player $1$ is trailing by $1$. We have the equations
$$\begin{align}a&=1+rb+(1-r)c,\\
b&=1+(1-r)a, \\
c&=1+ra.\end{align}$$
Use the above system of linear equations to find $a$.
Remarks: $1.$ To justify the first equation, note that for sure we will be playing one more game. That's the $1$ in the equation. And we will not be finished. With probability $r$ our expected number of games beyond that will be $b$, and with probability $1-r$ our expected number of games beyond that will be $c$. That yields the first equation. It may be prettier and clearer to write the equation as $a=r(1+b)+(1-r)(1+c)$.
The justifications for the other two equations are similar. Again, one might like to write the second equation as $b=r(1)+(1-r)(a+1)$, and do a similar rewrite of the third equation.
$2.$ As pointed out by Marc van Leeuwen, the argument tacitly assumes that the expectations exist. To show that they do is not difficult. Whatever $r$ is, the probability of two opposite results in a row is $2r(1-r)$, which is $\le 1/2$. So the probability that a game length $2n$ or more is $\le (1/2)^n$, and therefore the expected length is finite.
$3.$ We used a strategy broadly similar to the one used in the related question about the probability that Player $1$ wins. This kind of strategy tends to work more widely for expectations than for probabilities, because of the linearity of expectation.
Best Answer
With $p = 0.835, 2$ games are obviously not enough, because A would need to win both games, with a $Pr = {0.835}^2 <0.9$, so trials can begin from number of games played, $K = 3$
You would find computations much simpler if you focus on the losses of $A$ rather than wins.
For example, with $K=3$, $A$ can afford to lose at most one game, so the probability of an overall win is $\binom30q^0p^3 + \binom31q^1p^2$, which happily works out to
$\binom30 0.165^0\cdot 0.835^3 + \binom310.165^1\cdot0.835^2 = 0.9273$
thus the smallest value of n (games won) needed $=2$
Response to queries
We are trying to use the smallest number of games $(K)$ so we can find the smallest number of wins ($n$) needed.
With $K=3$, we found $n=2$ is possible if we include more than $2$ wins in the series of $K$ games.
If we want to strictly say that more than $n$ wins can't be included in the probability calculations, the formula will change to $\binom{K}{n} p^nq^{K-n} \geq 0.9$
A little thought will show that with this formula, the Pr will never reach $0.9$, even for $n=2$, see here
So either we'll have to allow more than $n$ wins in the Pr computations, else it is impossible.