There is a box with $n$ dice: $k$ of them are true, and $n - k$ unfair with $6$ on each one of the walls. We draw $m$ times (with replacement) one die and make a throw. Compute probability, that all drawn dice are fair provided that only $6$ were obtained (during these $m$ throws).
So:
$A\ - $ dice are fair
$B\ - $ only $6$ were thrown
$\Bbb P(A|B)=\frac{\Bbb P(A \cap B)}{\Bbb P(B)}$
where:
$\Bbb P(B) = \sum\limits_{i=0}^m{m \choose i}(\frac{k}{n})^i(\frac{n-k}{n})^{m-i}(\frac{1}{6})^i(\frac{6}{6})^{m-i}$
and $i$ is a number of fair dice;
$\Bbb P(A\cap B) = (\frac{k}{n})^m(\frac{1}{6})^m$
Am I correct?
While your work is correct, you can simplify $P(B)$. Note that the probability of $6$ in a throw for a randomly picked die is,
$p = \frac{k}{n} \cdot \frac{1}{6} + \frac{n-k}{n} \cdot 1 = 1 - \frac56 \cdot \frac k n$ $ \left(\frac56 \cdot \frac k n \text { is the probability of not getting $6$} \text { in a throw } \right)$
So probability of $m$ consecutive $6$ in $m$ throws is,
$P(B) = p^m = (1 - \frac56 \cdot \frac k n)^m $
The binomial expansion of this is same as the summation you wrote for $P(B)$.