Solution:

The answer is $\mathbf{(C)}$. Consider the first two characters of the keyword: since they must be different, we have $3 \cdot 2 = 6$ possibilities for choosing them. Without loss of generality, suppose they are $NE$. At this point we can carry out a simple case analysis, finding that if the third character is $V$ there are 4 possibilities, of which 2 also reach a length of 7 characters ($NEVEN V$, $NEVEN VN$, $NEVNV E$, $NEVNV EN$); moreover, none of these admits any further valid continuation. If instead the third character is $N$, the possibilities are only 2, one of 6 characters and one of 7 ($NENVEV$ and $NENVEVN$); indeed there exist some valid initial sequences that cannot be continued even up to 6 characters, such as for example $NENVN$.
At this point, for each of the 6 possibilities for the initial pair we have 6 (and always different among the various sets of 6, because indeed the initial pair distinguishes them), and so in total the possible keywords are $6 \cdot 6 = 6^{2}$.