Refer diagram.

We know that perimeter of a square is 4a and perimeter of a rectangle is 2(l+b).

Now let side of a square a = x+y.

Let c and d be sides of the rectangle.

To get value of c , we have a right triangle value of c = $\sqrt{2}$x.

Similarly value of d will be = $\sqrt{2}$y.

Now total perimeter of a rectangle will be 2( $\sqrt{2}$ x + $\sqrt{2}$ y ) .
=> 2($\sqrt{2}$(x+y)...

Now perimeter of square is 4(x+y).

Now question asks perimeter of sqaure what times greater than perimeter of rectangle.

4(x+y) = a times * 2($\sqrt{2}$)(x+y).

Cancel (x+y) on both sides.

4 = a times * 2($\sqrt{2}$) is possible when a is ($\sqrt{2}$)

So answer is A.

Hope it clears.