2020 AMC 10B Problem 7

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.