jmc

Let $a=x_1{x}_3{x}_5+x_2{x}_4{x}_6$ and $b=x_1{x}_2{x}_3+x_2{x}_3{x}_4+x_3{x}_4{x}_5+x_4{x}_5{x}_6+x_5{x}_6{x}_1+x_6{x}_1{x}_2.$ By AM-GM, $$a+b=(x_1+x_4)(x_2+x_5)(x_3+x_6)\le {\left[\frac{(x_1+x_4)+(x_2+x_5)+(x_3+x_6)}{3}\right]}^3=\frac{1}{27}.$$Hence, $$b\le \frac{1}{27}-\frac{1}{540}=\frac{19}{540}.$$Equality occurs if and only if $$x_1+x_4=x_2+x_5=x_3+x_6.$$We also want $a=\frac{1}{540}$ and $b=\frac{19}{540}.$ For example, we can take $x_1=x_3=\frac{3}{10},$ $x_5=\frac{1}{60},$ $x_2=\frac{1}{3}-x_5=\frac{19}{60},$ $x_4=\frac{1}{3}-x_1=\frac{1}{30},$ and $x_6=\frac{1}{3}-x_3=\frac{1}{30}.$

Thus, the maximum value of $b$ is $\boxed{\frac{19}{540}}.$