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Last digit of $196^{213}:$

We only need to care about the last digit. That is $6.$

We notice a repetition in the last digit of every $6^{n}$: it is $6.$

Now, last digit of $213^{196}:$

Similarily, we notice a repetition between $3^{1,2,3,4}:$ it is $3, 9, 7, 1.$

$196$ is a multiple of $4$ so the last digit is $1.$

$6 \cdot 1 = \boxed{6}.$