If the sum of the squares of first n natural numbers exceeds their sum by 330,

Sum of squares of first $n$ natural numbers is $\frac{n(n+1)(2 n+1)}{6}$

Sum of first $n$ natural number is $\frac{n(n+1)}{2}$

According to given condition, $\Sigma n^{2}-\Sigma n=330$

i. e $\frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}=330$

i. e $\frac{n(n+1)}{2}\left[\frac{2 n+1}{3}-1\right]=330$

i. e $\frac{n(n+1)}{2}\left[\frac{2 n+1-3}{3}\right]=330$

i. e $\frac{n(n+1)}{2}\left[\frac{2 n-2}{3}\right]=330$

i. e $\frac{n(n+1) 2(n-1)}{2 \times 3}=330$

i. e $n\left(n^{2}-1\right)=330 \times 3$

i. e $n^{3}-n=990$

i. e $n^{3}-n-990=0$

i.e n3 − 10n2 + 10n2 − n − 990 = 0 − 99n + 99n 

i.e n2 (n − 10) + 10n2 − 100n − 990 + 99n = 0

i.e n2 (n − 10) + 10n (n − 10) + 99(n−10) = 0

i.e (n − 10) (n2 + 10n + 99) = 0 

⇒ n = 10