= Solution
The graph has $\asymp n^4$ vertices, bounded degrees, and stationary masses $\asymp n^{-4}$. Split any set into components that do not communicate in one step and use part (a)(ii). A connected component of stationary mass $r\leq1/2$ has $O(rn^4)$ vertices. The planar grid isoperimetric bound supplies at least $c\sqrt{rn^4}$ boundary edges unless the component fills most of one layer; in that case the $\asymp n^2$ interlayer edges give the same order. Thus
$$
\Phi_*(r)\gtrsim\frac1{n^2\sqrt r}.
$$
The conductance-profile mixing bound for a lazy chain now gives
$$
t_{\mathrm{mix}}
\lesssim\int_{4\pi_{\min}}^{1/2}
\frac{du}{u\Phi_*(u)^2}+t_{\mathrm{rel}}
\lesssim n^4.
$$
Here $\Phi_*\gtrsim n^{-2}$ also gives $t_{\mathrm{rel}}\lesssim n^4$ by <Cheeger inequality>. Hence $t_{\mathrm{mix}}\lesssim n^4$.
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