jmc

Let the real, nonnegative roots be $u,$ $v,$ $w.$ Then by Vieta's formulas, $u+v+w=12$ and $uvw=64.$ By AM-GM, $$\frac{u+v+w}{3}\ge \sqrt[3]{uvw},$$which becomes $4\ge 4.$ This means we have equality in the AM-GM inequality. The only way this can occur is if $u=v=w,$ which means $u=v=w=4.$ Hence, the polynomial is $(x-4{)}^3=x^3-12x^2+48x-64,$ so $a=\boxed{48}.$