jmc

To start to eliminate the logarithms, we raise $8$ to the power of both sides, giving $$8^{{\log }_2({\log }_{8}x)}=8^{{\log }_8({\log }_{2}x)}$$or $$2^{3{\log }_2({\log }_{8}x)}=8^{{\log }_8({\log }_{2}x)},$$so $({\log }_{8}x{)}^3={\log }_{2}x.$ Now, by the change-of-base formula, ${\log }_{8}x=\frac{{\log }_{2}x}{{\log }_{2}8}=\frac{{\log }_{2}x}{3},$ so we have $${\left(\frac{{\log }_{2}x}{3}\right)}^3={\log }_{2}x.$$Thus $({\log }_{2}x{)}^2=3^3=\boxed{27}.$