The perimeter of square HJKL is 2 times the perimeter of square WXYZ
The 
SO if perimeter of HJKL is 2 times the perimeter the WXYZ than $$ \text{Perimeter of } HJKL = 2\cdot2(l+w)$$ So the L and W of $HJKL = 2l$ and $2w$? so the area is $2$ times larger than the $WXYZ$? But it looks like it was wrong.
$WYXZ$ has side length $a$ and perimeter $4a$.
$HJLK$ has side length $b$ and perimeter $4b = 2(4a) = 8 a$, so we know $b = 2 a$.
The area of $WYXZ$ is $a^2$, and $HJLK$ has area $b^2 = (2 a)^2 = 4 a^2$, thus four times the area of $WYXZ$.