Sum of squares of first n natural numbers exceeds their sum by 330, then $n=$

$\Sigma n=\frac{n\left(n+1\right)}{2}$

$Given\Sigma n^2=\Sigma n+330$

$\frac{n\left(n+1\right)\left(2n+1\right)}{6}=\frac{n\left(n+1\right)}{2}+330$

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

$n\left(n+1\right)\left(n-1\right)=3\times 330$

$\left(n+1\right)\left(n\right)\left(n-1\right)=11\times 10\times 9$

$n-1=9\Rightarrow n=10$