$\Pr[B\mid A]$ is the same as $\Pr[\text{$B$ given that $A$ has occurred}]$. Therefore, if you divide both sides of $\Pr[\text{$A$ and $B$}]=\Pr[A]\cdot\Pr[\text{$B$ given that $A$ has occurred}]$ by $\Pr[A]$, you get
$$
\Pr[B\mid A]=\frac{\Pr[\text{$A$ and $B$}]}{\Pr[A]},
$$
which is the conditional probability formula.
This can be used to solve your problem. Write
$$
\begin{aligned}
\Pr[\text{exactly 2 boys}\mid\text{at least 1 boy}]&=\frac{\Pr[\text{exactly 2 boys and at least 1 boy}]}{\Pr[\text{at least 1 boy}]}\\
&=\frac{\Pr[\text{exactly 2 boys}]}{\Pr[\text{at least 1 boy}]}.
\end{aligned}
$$
The second line follows from the first because there being exactly two boys implies that there is at least one boy.
If you treat the situation as a Bernoulli process, you can compute both of the needed probabilities.