If $X = 4$, there are two cases, namely, $A$ wins or $B$ wins. For the first case, $A$ must win in the $4$th set and win exactly two sets in the first three sets. Therefore,
$$
\Pr(X = 4, A \text{ wins}) = 0.6 \cdot \binom{3}{2}(0.6)^2(0.4)^1
$$
Similarly,
$$
\Pr(X = 4, B \text{ wins}) = 0.4 \cdot\binom{3}{2}(0.4)^2(0.6)
$$
The case of $X = 5$ is left as an exercise.
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.
Best Answer
Let $X$ be the random variable counting the number of bus drivers which would respond to a call on a particular day. If $4$ or more bus drivers respond positively, then all four buses will be operational and all children get picked up to go to school.
Let $n$ be the number of bus drivers (currently unknown).
$$.95\leq Pr(X\geq 4) = Pr(X=4 \text{ or }X=5 \text{ or } \cdots \text{ or } X=n)\\ = 1 - Pr(X=0\text{ or }X=1 \text{ or }X=2 \text{ or }X=3 \text{ or }X=4)\\ =1 - \binom{n}{0}.8^n + \binom{n}{1}.8^{n-1} .2^1 + \binom{n}{2} .8^{n-2}.2^2 + \binom{n}{3}.8^{n-3}.2^3$$
This can be graphed or solved via a computer program for an exact value of $n$ where equality with $.95$ is achieved. Else, you can guess and check different values of $n$ or consult a table to find the smallest such whole number $n$ such that the expression is greater or equal to $.95$.