jmc

Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$. If the perimeter of one is $3$ times the other's, then $a=3b$. The area of the larger square over the area of the smaller square is $$\frac{a^2}{b^2}=\frac{(3b{)}^2}{b^2}=\frac{9b^2}{b^2}=\boxed{9}$$