Figures $I$, $II$, and $III$ are squares.  The perimeter of $I$ is $12$ and the perimeter of $II$ is $24$.  The perimeter of $III$ is

$\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(D)}\ 81$
Since the perimeter of $I$ $12$, each side is $\frac{12}{4} = 3$.  
Since the perimeter of $II$ is $24$, each side is $\frac{24}{4} = 6$.
The side of $III$ is equal to the sum of the sides of $I$ and $II$.  Therefore, the side of $III$ is $3 + 6 = 9$.
Since $III$ is also a square, it has an perimeter of $9\cdot 4 = 36$, and the answer is $\boxed{C}$.