smc

Since the diagonal of square $I$ is $a+b$ units long, the side length of square $I$ is $\frac{a+b}{\sqrt{2}}$, so the area of square $I$ is $\frac{(a+b{)}^2}{2}$. The area of square $II$ is twice as much as the area of square $I$, so the area of square $II$ is $(a+b{)}^2$. That means the side length of square $II$ is $a+b$, so the perimeter of square $II$ is $4(a+b)$, or answer choice $\boxed{4(a+b)}$.