smc

The area of this square is equal to $6+8+30+14+6=64$, and thus its side lengths are $8$. The sum of the dimensions of the rectangles are $2+7+5+6+2+3+1+6+2+4=38$. Thus, because the perimeter of the rectangle is $32$, the rectangle on the inside must have a perimeter of $6\cdot 2=12$. The only rectangle that works is $\boxed{B}$.