 Solution

There are $\asymp n^3$ edges. The $n^2$ horizontal cutsets between successive rows perpendicular to the long direction are disjoint and each contains $\asymp n$ unit-conductance edges. Nash-Williams inequality givesConversely, spread a unit flow nearly uniformly across the width $n$ while moving it through the $n^2$ long-direction layers, with bounded extra energy to fan out from and collect at the two corner vertices. Its energy is $O(n^2/n)+O(\log n)=O(n)$, so Thomson's principle gives the matching upper bound $R_{\mathrm{eff}}\asymp n$.

The graph is invariant under a half-turn exchanging the two corners, so the two directional hitting-time expectations are equal. The commute time identity therefore yields