Exercise 6.3 - $\hat{\sigma}^{2}_{\text{MLE}}$ is biased

Consider:

The second term on the r.h.s. contains $N$ components that equal to $𝔼[x^2]={\mu }^2+{\sigma }^2$ and $N^2-N$ components that equal to $𝔼[x_l{x}_m]={\mu }^2$, where $l\ne m$. For the third term, the numbers are $1$ and $N-1$, thus:

so is ${\hat{\sigma }}_{\text{MLE}}^2$. The unbiased estimation, of which one who has taken several elementary courses on probability and statistics should have been very familiar, is: