Let the length of a short side of a rectangle be $x$ and the length
of a long side be $y$. Then the whole square has side of length
$(y+x)$, whilst the small square has side of length $y-x$. As the
area of the whole square is four times the area of the small
square, the length of the side of the whole square is twice the
length of the side of the small square. Therefore $y+x=2(y-x)$,
i.e., $y=3x$ so $x:y=$1:3.
This problem is taken from the UKMT Mathematical Challenges.