$r_n=n^2\times r_1$

$\Rightarrow r_{n+1}=(n+1)^2\times r_1$

$\Rightarrow r_{n-1}=(n-1)^2\times r_1$

$\Rightarrow r_{(n+1)}-r_{n}=r_{(n-1)}$

$\therefore (n+1)^2r_1-n^2r_1=(n-1)^2r_1$

$\Rightarrow (n^2+2n+1)-n^2=(n^2-2n+1)$

$\Rightarrow n^2-4n=0$

$\therefore n=4$

Hence (c) is the correct answer.