i loop is executing $\Theta(n)$ times.
j loop is executing $\Theta(\log n)$ times for each i. (So, time complexity must for code must be $\Omega(n \log n)$
But it is not straight forward to count the no of iterations of k loop as it depends on the value of j. So, we can see the values of j which goes like 1, 2, 4, ... n. So, no. of iterations of k loop (over all j values) will be $\sum_{j=0}^{\lg n} 2^j = 2^{\lg n +1} = \Theta(n)$. And this is over all j loops (not including i). Including i loop we get time complexity as $\Theta\left(n^2 \right).$