According to the question:
$\frac{\text{area of 1st square}}{\text{area of 2nd square}}=\frac{9}{1}$
$\because$ Area of square is $a^2$,where $a$ is side of square.
Assume that side of square1 is $a_1$ & side of square 2 is $a_2$.
$\implies$ $\frac{(a_1)^2}{(a_2)^2}=\frac{9}{1}$
$\implies$ $\frac{a_1}{a_2}=\frac{3}{1}$
$\therefore$ Side of square1 is $3$ and side of square2 is $1$
$\because$ perimeter of square is $4*a$,so there ratio is;
$\implies$ $\frac{4*a_1}{4*a_2}$
$\implies$ $\frac{4*3}{4*1}$
$\implies$ $3:1$
Option (b) is correct.
