More generally, one can count the number of $2$ pairs in a $5$ card poker hands with $$ {13\choose 2}{4\choose 2}^2{11\choose 1}{4\choose 1}. $$
The hand you describe is on of a specific subsets hand in these hands, so we have to count them. You can choose $2$ of $4$ kings, $2$ of $4$ aces, and $1$ of $4$ queen. This give $$ {4\choose2}{4\choose2}{4\choose 1}=144. $$ We can calculate your probability as follow: Define $T$ the event having two pairs and $H$ the event having the hand you specify. Then $$ \mathbb{P}(H\cap T)=\mathbb{P}(H|T)\mathbb{P}(T) $$ But $H\cap T$ is just $H$ so $$ \mathbb{P}(H)=\frac{144}{{13\choose 2}{4\choose 2}^2{11\choose 1}{4\choose 1}}\frac{{13\choose 2}{4\choose 2}^2{11\choose 1}{4\choose 1}}{{52\choose 5}}=\frac{144}{{52\choose 5}}=\frac{3}{54145} $$
This seems unnecessarly complicated but uses many theorems/property of probability along with combinatorial arguments and shows you that your second method was right.