 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 $r\le 1/2$ has $O(r{n}^4)$ vertices. The planar grid isoperimetric bound supplies at least $c\sqrt{r{n}^4}$ boundary edges unless the component fills most of one layer; in that case the $\asymp n^2$ interlayer edges give the same order. ThusThe conductance-profile mixing bound for a lazy chain now givesHere ${\Phi }_{\ast }\gtrsim n^{-2}$ also gives $t_{\mathrm{rel}}\lesssim n^4$ by Cheeger inequality. Hence $t_{\mathrm{mix}}\lesssim n^4$.