Let $V_i$ to be the set of vertices with distance exactly $i$ from $s$. The maximum flow is clearly bounded by $|V_i||V_{i+1}|$, as all flows has to go in between the two levels and the maximum number of edges in between happens when it is a complete bipartite graph. Notice in order to maximize this value, we have $|V_i|=n/d$. Hence $v\le 2(n/d{)}^2$. Similarly, we can also distribute the number of edges in between every two partitions, and get $v\le 2m/d$.

$v\le 2(n/d{)}^2$, thus $d\le n\sqrt{2}/\sqrt{v}$. If we take the shortest path successively, we have the number of edges we use in each augmenting path is

$$\sum _{i=1}^{v}n\sqrt{2}/\sqrt{i}=O(n\sqrt{v})$$

Let $f$ be an acyclic flow of value $v$ in $G$. Remove all edges outside $f$, and let the remaining graph be $G^{\prime }$. Consider use shortest augmenting path algorithm sending a flow of value $v$ in $G^{\prime }$. Notice this would produce $f$. Thus this shows $f$ has $O(n\sqrt{v})$ edges.

There exist $\lambda (s,t)$ $st$-edge disjoint paths with total length $O(\sqrt{\lambda (s,t)}n)$. The average length is $O(\frac{n}{\sqrt{\lambda (s,t)}})$. The $k$ shortest paths has total length $O(k\frac{n}{\sqrt{\lambda (s,t)}})=O(k\frac{n}{\sqrt{k}})=O(\sqrt{k}n)$.