jmc

By AM-GM, $$abc=a+b+c\ge 3\sqrt[3]{abc},$$so $(abc{)}^3\ge 27abc,$ which means $(abc{)}^2\ge 27.$

Since $bc=a^2,$ $a^6\ge 27,$ so $a^2\ge 3.$

Equality occurs when $a=b=c=\sqrt{3},$ so the smallest possible value of $a^2$ is $\boxed{3}.$