The answer should be option b.
The basic operations of Dijkstra's algorithm are extract-min and Relax(decrease key).
Now for sorted linked list extract-min would take $O(1)$ and Relax operation would take $O(V)$. And as we know the Relax operation will be applied on every edge so upper bound would become $O(VE)$.
Now in worst case $E=|V|^2$ , so time complexity $= O(V^2.V) =O(|V|^3).$