There are $14$ non-red balls, so there are $\binom{14}5$ ways to draw $5$ non-red balls. There are $\binom{20}5$ ways to draw $5$ balls regardless of color, so there are $\binom{20}5-\binom{14}5$ ways to draw $5$ balls including at least one red ball.
Now we’ll calculate the number of ways to draw $5$ balls, of which at most one is green. The easiest way to calculate this is to split it into two cases: no green balls, and exactly one green ball. Since there are $6$ green balls, there are $14$ non-green balls, and the number of ways of drawing $5$ of them is $\binom{14}5$, just as when we counted the ways of drawing $5$ non-red balls. There are $\binom61=6$ ways to draw one green ball and $\binom{14}4$ ways to draw $4$ non-green balls, so there are $6\binom{14}4$ ways to draw $5$ balls, exactly one of which is green. Adding the two cases, we see that there are $\binom{14}5+6\binom{14}4$ ways to get at least one green ball amongst the $5$.
The final answer, as you realized, is the product of these two numbers,
$\left(\binom{20}5-\binom{14}5\right)\left(\binom{14}5+6\binom{14}4\right)\;.$
I make that $(15504-2002)(2002+6006)=108,124,016$.