A 3x3x3 cube is made of $27$ normal dice. Each die's opposite sides sum to $7$. What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?
$\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96$
In a 3x3x3 cube, there are $8$ cubes with three faces showing, $12$ with two faces showing and $6$ with one face showing. The smallest sum with three faces showing is $1+2+3=6$, with two faces showing is $1+2=3$, and with one face showing is $1$. Hence, the smallest possible sum is $8(6)+12(3)+6(1)=48+36+6=90$. Our answer is thus $\boxed{\text{(D)}\ 90}$.