Calculating the Time complexity using Radix sort
You are correct.  We have $(n/\log n)^{20} \le n^{10}$ for all $n\ge 1$, so each number can be expressed using at most $d=20$ &quot;digits&quot;.  The running time of radix sort is $O(dn)$, and $O(20n)$ is $O(n)$, so the running time is linear in $n$.