ABCs

Since $\begin{array}{rcl} (m^2 + 4n^2)^3 & = & m^6 + 12m^4n^2 + 48m^2n^4 + 64n^6 \; = \; (m^6 - 24m^4n^2 + 144m^2n^4) + 4(9m^4n^2 - 24m^2n^4 + 16n^6) \\ & = & (m^3 - 12mn^2)^2 + 4(3m^2n - 4n^3)^2 \; = \; 1600 + 4\times100 \; = \; 2000 \end{array}$ we have $m^2 = 4n^2 \,=\, \boxed{10\sqrt[3]{2}}$ .