A more probabilistic version of the idea used below, indeed essentially a full solution, was given in a comment by @Srivatsan Narayanan.
Conveniently, the odd digits have mean $3$, and so do the even digits.
So we only need to count the number of allowed permutations, and multiply by $3$. It is somewhat easier, or at least safer, to count the complement.
There are $(3!)(2)$, that is, $12$ permutations with all the odd digits in odd positions.
Next we count the permutations with exactly one odd digit in an odd position. That digit can be chosen in $3$ ways. For each choice, its position can be chosen in $3$ ways. Once we have done that, the remaining odd digits can be placed in $2$ ways, and then the even digits can be placed in $2$ ways, for a total of $(3)(3)(2)(2)$, that is, $36$.
So there are $48$ forbidden permutations, leaving $5!-48=72$ allowed permutations.
The required total is therefore $(3)(72)$, which is $216$.