We assume that balls of the same colour are identical. Let $r$ be the number of red we pick, $g$ the number of green, and $b$ the number of blue. We want to find the number of solutions of the equation $r+g+b=4$ in non-negative integers.
This would be taken care of by a general formula ("Stars and Bars") if there were $4$ or more balls in each bucket. However, there are only $2$ blue balls, so for example the solution $(0,1,3)$ is not possible.
I do not know whether you are expected to use a "general" procedure, or just plain count. We will do the latter, in a reasonably organized manner.
Divide into cases.
4 red: There is only $1$ way to do this.
3 red: We can pick a green, or a blue, $2$ ways.
2 red: We pick $2$ green, or $2$ blue, or $1$ of each, a total of $3$ ways.
1 red: We pick $3$ green, or $2$, or $1$, and the rest blue, a total of $3$ ways.
0 red: We pick $3$ green, or $2$, and the rest blue, a total of $2$ ways.
Remark: Since the blues are fewest, it is obviously more efficient to let the cases be $2$ blue, $1$ blue, $0$ blue. You should do it that way, in order to make the solution "your own."