The sum of the squares of the first n odd natural number is given by;
$S_n=\frac{n(2n+1)(2n-1)}{3}$
It is given that $S_n=533n$ so;
$\implies533n*3=n((2n)^2-1)$
$\implies 1599*n=4n^3-n$
$\implies 1599*n=n(4n^2-1)$
$\implies 1599+1=4n^2$
$\implies 1600=4n^2$
$\implies n^2=\frac{1600}{4}$
$\implies n^2=400$
$\implies n=20$
Option (B) is correct.