Let the sides of the two squares be $a$ m and $b$ m ($a>b$). Given: $a^2+b^2=640$ ...(i). Difference of perimeters: $4a-4b=64 \Rightarrow a-b=16 \Rightarrow a=b+16$ ...(ii). Substitute (ii) in (i): $(b+16)^2+b^2=640$. $b^2+32b+256+b^2=640$. $2b^2+32b-384=0$. $b^2+16b-192=0$. Using the quadratic formula: $b=\frac{-16\pm\sqrt{256+768}}{2}=\frac{-16\pm\sqrt{1024}}{2}=\frac{-16\pm32}{2}$. $b=8$ (rejecting negative value). $a=b+16=24$. Sides of the two squares are 24 m and 8 m.