Sum of squares of first n natural numbers = $\dfrac{n × (n + 1) (2n + 1)}{6}$
Try estimating the options by picking values for n:
n = 10; [n(n+1)(2n+1)]/6 = 385
n = 11; [n(n+1)(2n+1)]/6 = 506
Notice that the option (a) 420 falls in between the two values 385 (n = 10) and 506 (n = 11). This means, it is not a sum of squares of first n natural numbers. Hence this is the answer.