Solution:

A number that is both even and a multiple of $3$ is a multiple of $6$. If such a number is also a perfect square, its square root must be divisible by both $2$ and $3$, hence by $6$. Therefore the numbers counted are exactly $$(6k)^2<2020$$ for positive integers $k$.

Since $42^2=1764<2020$ and $48^2=2304>2020$, we have $k=1,2,\ldots,7$, for a total of $7$ numbers.

Thus, A is the correct answer.