jmc

Expanding, we get $${\left(2a+\frac{1}{3b}\right)}^2+{\left(2b+\frac{1}{3c}\right)}^2+{\left(2c+\frac{1}{3a}\right)}^2=4a^2+\frac{4a}{3b}+\frac{1}{9c^2}+4b^2+\frac{4b}{3c}+\frac{1}{9c^2}+4c^2+\frac{4c}{3a}+\frac{1}{9a^2}.$$By AM-GM, $$4a^2+\frac{1}{9c^2}+4b^2+\frac{1}{9c^2}+4c^2+\frac{1}{9a^2}\ge 6\sqrt[6]{4a^2\cdot \frac{1}{9c^2}\cdot 4b^2\cdot \frac{1}{9c^2}\cdot 4c^2\cdot \frac{1}{9a^2}}=4$$and $$\frac{4a}{3b}+\frac{4b}{3c}+\frac{4c}{3a}\ge 3\sqrt[3]{\frac{4a}{3b}\cdot \frac{4b}{3c}\cdot \frac{4c}{3a}}=4.$$Hence, $$4a^2+\frac{4a}{3b}+\frac{1}{9c^2}+4b^2+\frac{4b}{3c}+\frac{1}{9c^2}+4c^2+\frac{4c}{3a}+\frac{1}{9a^2}\ge 8.$$Equality occurs when $2a=2b=2c=\frac{1}{3a}=\frac{1}{3b}=\frac{1}{3c}$ and $\frac{4a}{3b}=\frac{4b}{3c}=\frac{4c}{3a},$ or $a=b=c=\frac{1}{\sqrt{6}},$ so the minimum value is $\boxed{8}.$