jmc

Let the integer roots be $r,$ $r,$ and $s,$ so $$x^3+a{x}^2+bx+9a=(x-r{)}^2(x-s).$$Expanding and matching coefficients, we get $$\begin{array}{rl}2r+s& =-a,\\ r^2+2rs& =b,\\ r^{2}s& =-9a.\end{array}$$From the first and third equations, $r^{2}s=9(2r+s),$ so $$s{r}^2-18r-9s=0.$$As a quadratic in $r,$ the discriminant is $$\sqrt{18^2-4(s)(-9s)}=\sqrt{324+36s^2}=3\sqrt{s^2+9}.$$Since $r$ and $s$ are integers, $s^2+9$ must be a perfect square. Let $s^2+9=d^2,$ where $d>0.$ Then $$(d+s)(d-s)=9.$$If $s=0,$ then $a=0,$ which is not allowed. Otherwise, $d=\pm 5$ and $s=\pm 4.$ If $s=4,$ then $r=6,$ and $a=-16$ and $b=84.$ If $s=-4,$ then $r=-6,$ and $a=16$ and $b=84.$ In either case, $$|ab|=16\cdot 84=\boxed{1344}.$$