The answer is $w_k = \lambda_k/\mu$ and $w_i = \lambda_i$ for $i \neq k$.
Multiplying the $k$th constraint of the primal by $\mu$ doesn't change the primal feasible region and so doesn't change the primal optimal solution. Thus the value $z^*$ of the optimal primal solution doesn't change.
The only change to the dual problem is that the $k$th variable is replaced by $\mu$ times that variable (in the constraints and the objective). Since $z^*$ is the value of the original dual optimal solution, letting $w_i = \lambda_i$ for $i \neq k$ and $w_k = \lambda_k/\mu$ yields a solution that is feasible for the new dual and that also has value $z^*$. Thus, by weak duality, $w$ must be optimal.