I assume that you mean $z^3 -6z^2 + 13z + k = 0$

I also assume you mean $k$ is real. Since all the coefficients are real you know that roots come in conjugate pairs. Or you would if you read your textbook.

Thus you have two of the roots, $2+i, ~ 2-i$.

Divide your polynomial by $(x-(2+i))(x-(2-i))$ and you'll end up with a first degree polynomial quotient with an obvious zero and a remainder that must be zeroed out.

Zeroing out the remainder allows you to solve for $k$