Assume you have a simple, infinite graph $G$ with bounded degree (there is an upper bound for the degree of the nodes). Choose an arbitrary vertex $x\in V(G)$ and consider $$ G_{n}:=\{x\in G:d(x_0,x)\leq n\} $$ with the graph metric (hop metric). Assume that each pair of nodes is communicating a unit load of information and the load goes through the minimum path between nodes (if there is more than one minimum path we choose one arbitrarily). The total traffic in $G_{n}$ is equal to $\frac{N(N-1)}{2}$ where $N=N(n)=|G_{n}|$.
Given a node $v\in G_{n}$ we define $T_{n}(v)$ as the total traffic generated in $G_{n}$ passing through $v$. In other words, $T_{n}(v)$ is the sum off all the geodesic paths in $G_{n}$ which are carrying traffic and contain the node $v$.
If the graph $G$ is planar and it has exponential growth $|G_{n}|=K\exp(\lambda n)$ for $n$ sufficiently large, then it is not difficult to prove that there are nodes in $G_{n}$ such that $$ T_{n}(v)\geq C\frac{N^2}{\log(N)} $$ for $n$ sufficiently large.
My question is the following
Is the same true if we remove the planar condition but we keep the exponential growth? My intuition is that the answer is no but I can't find a counterexample.