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.