jmc

Inverting each logarithm, we have $$\frac{1}{{\log }_{10}10x^2}+\frac{1}{{\log }_{10}100x^3}=-2,$$or $$\frac{1}{1+2{\log }_{10}x}+\frac{1}{2+3{\log }_{10}x}=-2.$$Now, make the substitution $y={\log }_{10}x,$ giving $$\frac{1}{1+2y}+\frac{1}{2+3y}=-2.$$To solve this equation, we multiply both sides by $(1+2y)(2+3y)$ to get $$(2+3y)+(1+2y)=-2(1+2y)(2+3y),$$which rearranges to $$12y^2+19y+7=0.$$Factoring this quadratic, we get $$(y+1)(12y+7)=0,$$so either $y=-1$ or $y=-\frac{7}{12}.$ Since $y={\log }_{10}x,$ we have $x=10^y,$ so either $x=10^{-1}$ or $x=10^{-7/12}.$ The larger of these two solutions is $x=10^{-7/12},$ so the answer is $$\frac{1}{x^{12}}=x^{-12}=10^7=\boxed{10000000}.$$