The outer loop executes $\Theta(\log n)$ time as $k$ goes from $n, n/2, \dots 0$. The inner loop is bit tricky.
The $j$ values goes like
$1, 1+2, 1+2+3, 1+2+3+4, \dots n$
This means j loop iterates $\Theta(\sqrt (n)) $ times as it eventually stops when the sum of first $k$ integers $(\frac{k.(k+1)}{2})$ goes above $n$.
So, overall time complexity is $\Theta \left( \log n \sqrt n \right)$