= 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> gives
$$
R_{\mathrm{eff}}((0,0),(n,n^2))\gtrsim n^2/n=n.
$$
Conversely, 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
$$
\mathbb E_{(0,0)}T_{(n,n^2)}
\asymp |E|R_{\mathrm{eff}}\asymp n^3n=n^4.
$$
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