jmc

By Vieta's formulas, $rs+st+tr=18.$ Squaring this equation gives us the terms we want: $$(rs+st+tr{)}^2=(rs{)}^2+(st{)}^2+(tr{)}^2+(2r^{2}st+2r{s}^{2}t+2rs{t}^2)=324.$$To deal with the extra terms, we note that $$r^{2}st+r{s}^{2}t+rs{t}^2=rst(r+s+t)=7\cdot 20=140,$$again by Vieta. Therefore, $$(rs{)}^2+(st{)}^2+(tr{)}^2=324-2\cdot 140=\boxed{44}.$$