Outermost loop having $i$ as iterator executes $O(n)$ time. (No doubt here)
Let's decode other two loops $\Rightarrow$
for(j=1;j<n;j*=2) {
for(k=0;k<j;k++)
}
When $j= 1$, Innermost loop executes 1-time.
When $j = 2$, Innermost loop executes 2-times.
When $j = 4$, Innermost loop exexutes 4-times.
So, total number of times innermost loop executes $= 1+2+4 + \dots + 2^k = \frac{2^{k+1} -1}{2-1} = 2^{k+1} -1$
And condition j < n evaluates to be false when $2^{k+1}-1 \ge n$, thus $k = O(\log_2n)$
So, Overall time complexity $= O(n) * O(2^{k+1}-1) = n*O(2^{\log_2n}) = O(n^2)$
