Probability of exactly $2$ sixes in $3$ dice rolls where $2$ dice have $6$ on $2$ faces?

Note that the official answer is correct if you make the (unnatural) assumption that the two non-standard dice have seven sides (two of which show $6$).

In that case the answer is $$\frac 27\times \frac 27\times \frac 56+2\times \frac 27\times \frac 57\times \frac 16=\frac {20}{147}$$

To be sure, this was arrived at by reverse engineering, not by any sensible reading of the problem.

Your solution seems correct, indeed also by the naif definition of probability we obtain

$$p=\frac{\text{#favorauble cases}}{\text{#total cases}}=\frac{8+8+20}{6^3}=\frac16$$