In the product shown, $\text{B}$ is a digit.  The value of $\text{B}$ is
\[\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$
Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.
So, $2 \times \text{B}$ has a units digit of $6$, so $\text{B}$ is either $3$ or $8$. If $\text{B}=3$, then the product is $32\times 73$, which is clearly too small, so $\text{B}=8$
$\boxed{\text{E}}$