 Solution

A complete excursion from $B$ along any one arm and back to $B$ has mean length $L=2n$. This follows from the same path recurrence, or from Kac's lemma: the graph has $3n$ edges, the stationary distribution of simple random walk assigns mass $\mathrm{deg}(B)/(2|E|)=3/(6n)=1/(2n)$ to $B$, so its mean return time is $L=2n$.

Let $x$ be the mean time to hit $E$ starting at $B$. On each visit to $B$, the first step reaches $E$ with probability $1/3$; with probability $2/3$ it begins a wrong-arm excursion of mean duration $L$, after which the problem restarts. ThusThe walk first takes mean time $n^2$ to reach $B$, so the requested time from $A$ is