jmc

To compare different values of $A_k,$ we look at the ratio $A_k/A_{k-1},$ which equals $$\frac{A_k}{A_{k-1}}=\frac{\left(\genfrac{}{}{0ex}{}{1000}{k}\right)(0.2{)}^k}{\left(\genfrac{}{}{0ex}{}{1000}{k-1}\right)(0.2{)}^{k-1}}=\frac{\frac{1000!}{k!(1000-k)!}(0.2{)}^k}{\frac{1000!}{(k-1)!(1001-k)!}(0.2{)}^{k-1}}=\frac{1001-k}{5k}.$$The inequality $$\frac{A_k}{A_{k-1}}=\frac{1001-k}{5k}>1$$holds if and only if $k<\frac{1001}{6}=166.8\overline{3},$ that is, if $k\le 166.$ Therefore, $A_k>A_{k-1}$ holds when $k\le 166,$ and $A_k<A_{k-1}$ holds when $k\ge 167.$ Thus, $$A_{166}>A_{165}>\cdots >A_1$$and $$A_{1000}<A_{999}<\cdots <A_{166},$$which means that $A_k$ is largest for $k=\boxed{166}.$