$A_1=\{2k: k\in \mathbb{N}\}=\{2,4,6,8,10,12,14,16,18,...,2k,...\}$
$A_2=\{3k: k\in \mathbb{N}\}=\{3,6,9,12,15,18,21,24,27,...,3k,...\}$
$\begin{align} \therefore A_1 \cap A_2&=\{6,12,18,...,6k,...\}\\&=\{6k:k \in \mathbb{N}\}\\&=\{(5+1)k:k \in \mathbb{N}\}\\&=A_5\end{align}$
So the correct answer is C.