The log-likelihood of a Gaussian sample $X_1 \dots X_n$ with $X_i\sim \mathcal{N}(\mu_0,\sigma^2)$ is
\begin{equation}
L = -\frac{n}{2} \ln(2\pi) - \frac{n}{2} \ln\sigma^2 - \frac{1}{2\sigma^2} \sum_{i=1}^n \left(X_i - \mu_0\right)^2 \, ,
\end{equation}
where $\sigma^2$ is the parameter to estimate.
The derivative of the log-likelihood with respect to $\sigma^2$ vanishes at $\sigma^2 = \hat{\sigma}^2_n$, with
\begin{equation}
\hat{\sigma}^2_n = \frac{1}{n} \sum_{i=1}^n \left(X_i - \mu_0\right)^2 \, .
\end{equation}
One shows that $\sigma^2 = \hat{\sigma}^2_n$ corresponds to a maximum of the log-likelihood. Therefore, $\hat{\sigma}^2_n$ defines the maximum likelihood estimator. Now, we compute the expected value of the estimator
\begin{equation}
E(\hat{\sigma}^2_n) = \frac{1}{n} \sum_{i=1}^n E\left(\left(X_i - \mu_0\right)^2\right) .
\end{equation}
Since $\mu_0 = E(X_i)$, one has $E\left(\left(X_i - \mu_0\right)^2\right) = \sigma^2$ for all $i$. Finally, $E(\hat{\sigma}^2_n) = \sigma^2$. The estimator is unbiased.
For comparison, in the case where the mean is unknown, the log-likelihood
\begin{equation}
L = -\frac{n}{2} \ln(2\pi) - \frac{n}{2} \ln\sigma^2 - \frac{1}{2\sigma^2} \sum_{i=1}^n \left(X_i - \mu\right)^2
\end{equation}
is maximized with respect to the parameter $(\mu,\sigma^2)$. The maximum likelihood estimator is
\begin{aligned}
&\tilde{\mu}_n = \overline{X}_n = \frac{1}{n} \sum_{i=1}^n X_i\, ,\\
&\tilde{\sigma}^2_n = \frac{1}{n} \sum_{i=1}^n \left(X_i - \overline{X}_n\right)^2\, ,
\end{aligned}
where the variance estimator $\tilde{\sigma}^2_n$ is biased: $E(\tilde{\sigma}^2_n) = \frac{n-1}{n}\sigma^2 \neq \sigma^2$.
