jmc

Let $r,$ $s,$ $t$ be the real roots, so $$r^3-a{r}^2+br-a=0.$$If $r$ is negative, then $r^3,$ $-a{r}^2,$ $br,$ and $-a$ are all negative, so $$r^3-a{r}^2+br-a<0,$$contradiction. Also, $r\ne 0,$ so $r$ is positive. Similarly, $s$ and $t$ are positive.

By Vieta's formulas, $r+s+t=a$ and $rst=a.$ By AM-GM, $$\frac{r+s+t}{3}\ge \sqrt[3]{rst}.$$Then $$\frac{a}{3}\ge \sqrt[3]{a}.$$Hence, $a\ge 3\sqrt[3]{a},$ so $a^3\ge 27a.$ Since $a$ is positive, $a^2\ge 27,$ so $a\ge 3\sqrt{3}.$

Equality occurs if and only if $r=s=t=\sqrt{3},$ so the cubic is $$(x-\sqrt{3}{)}^3=x^3-3x^2\sqrt{3}+9x-3\sqrt{3}=0.$$Thus, $b=\boxed{9}.$