Being $k$-rough is the same as being relatively prime to the product of primes up to $k$. Hence if $P(k)$ denotes that product, and $j$ denotes the [Jacobsthal function][1], then $t=j(P(k))-1$ works for every $n$, while $t=j(P(k))-2$ fails for infinitely many $n$'s. In particular, it follows from a result of [Iwaniec (1978)][2] that $t=ck^2$ works for some absolute constant $c>0$.


  [1]: https://oeis.org/wiki/Jacobsthal_function
  [2]: https://www.degruyter.com/document/doi/10.1515/dema-1978-0121/html