understand the solution to the unbiased estimator of area of circle when given n independent radius $R$ measurement with error $\sim N(0,\sigma^2)$ $S =  \pi R^2$
$E(\bar{X}^2) = Var(\bar{X})+(E(\bar{X}))^2=\sigma^2/n + R^2$
then it states thats an unbiased estimator of $\sigma^2/n = \frac{1}{n(n-1)} \sum_{i=1}^n(X_i-\bar{X})^2$.
However, when i approached it by finding the unbiased estimator of $\sigma^2$ in $N(\bar{X},\sigma^2)$ from its MLE, the result is $\frac{n}{n-1}(X_i-\bar{X})^2$. Thus I don't understand where the extra division by $n$ is coming from.
Suppose $R_1,\ldots,R_n\sim N(R,\sigma^2)$ are $n$ independent measurements of your radius $R$. Set $A_i=\pi R_i^2$ and $S^2=\frac{1}{n-1}\sum_{i=1}^n(R_i-\bar{R})^2$. Notice that $S^2$ is an unbiased estimator for $\sigma^2$ and $$E(A_i)=\pi E(R_i^2)=\pi \Big[\sigma^2+\big(E(R_i)\big)^2\Big]=\pi E(S^2) + \pi R^2$$ If we take $\bar{A}=\frac{1}{n}\sum_{i=1}^nA_i$ it's easy to see that $E(\bar{A})$ also equals $\pi E(S^2)+\pi R^2$ and so $\hat{A}=\bar{A}-\pi S^2$ is an unbiased estimator for $A=\pi R^2$.
Remark: If you want to know why $S^2$ is an unbiased estimator for $\sigma^2$, note $\frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}$. Hence $$E\Big(\frac{(n-1)S^2}{\sigma^2}\Big)=n-1 \iff E(S^2)=\sigma^2$$
Suppose $R_1,\ldots,R_n\sim N(R,\sigma^2)$ are $n$ independent measurements of your radius $R$. Set $A_i=\pi R_i^2$ and $S^2=\frac{1}{n-1}\sum_{i=1}^n(R_i-\bar{R})^2$. Notice that $S^2$ is an unbiased estimator for $\sigma^2$ and $$E(A_i)=\pi E(R_i^2)=\pi \Big[\sigma^2+\big(E(R_i)\big)^2\Big]=\pi E(S^2) + \pi R^2$$ If we take $\bar{A}=\frac{1}{n}\sum_{i=1}^nA_i$ it's easy to see that $E(\bar{A})$ also equals $\pi E(S^2)+\pi R^2$ and so $\hat{A}=\bar{A}-\pi S^2$ is an unbiased estimator for $A=\pi R^2$.
Remark: If you want to know why $S^2$ is an unbiased estimator for $\sigma^2$, note $\frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}$. Hence $$E\Big(\frac{(n-1)S^2}{\sigma^2}\Big)=n-1 \iff E(S^2)=\sigma^2$$