In [1, p. 61 exercise 3.5] we are asked to show that for the conditions of Problem [1, p. 60, exercise 3.4] the CR bound is
var(\hat{\mu}_x) \geq \frac{\sigma_x^2}{N}

and further to give a statement about the efficiency of the sample mean estimator.
Solution: The Cramer Rao bound of the variance of an estimated parameter \hat{\vartheta} of a propability density function p(\mathbf{x},\vartheta) which is obtained from samples of the random variables \mathbf{x} is obtained by the formula [1, p. 46, (3.17),(3.18)] :
var(\hat{\vartheta}) \geq \frac{1}{-E\left\{\frac{\partial^2 \ln{p(\mathbf{x};\vartheta)}}{\partial \vartheta^2}\right\}} (1)

Because the samples {x[0],…,x[n-1]} are independent random variables with normal distribution N(\mu_x, \sigma_x^2) the joint propability density function is given by:
f(\mathbf{x}) =\prod_{i=0}^{N-1}\left(\frac{1}{\sqrt{2\pi}\sigma_x}e^{-\frac{1}{2}\left(\frac{x[i]-\mu_x}{\sigma_x}\right)}\right) (2)

thus:
\ln{f(\mathbf{x})}=-\ln(\sqrt{2\pi}\sigma_x) N-\frac{1}{2}\sum\limits_{i=0}^{N-1}\left(\frac{x[i]-\mu_x}{\sigma_x}\right)^2 \Rightarrow (3)
\frac{\partial\ln{f(\mathbf{x})}}{\partial\mu_x}=\sum\limits_{i=0}^{N-1}\left(\frac{x[i]-\mu_x}{\sigma_x^2}\right) \Rightarrow
\frac{\partial^2\ln{f(\mathbf{x})}}{\partial\mu_x^2} = \frac{N}{\sigma_x^2}

From the previous relation and (1) we obtain the result that the Cramer Rao bound of the variance of the mean estimation is bounded by \frac{\sigma_x^2}{N}:
\sigma_{\hat{\mu}_x} \geq \frac{\sigma_x^2}{N} (4)

We had computed in [2, relation (5)] that the variance of the estimator of the mean \hat{\mu}_x=\frac{1}{N}\sum_{i=0}^{N-1}x[i] is given by var(\hat{\mu}_x)=\frac{1}{N}\sigma_x^2. Thus the sample mean \hat{\mu}_x=\frac{1}{N}\sum_{i=0}^{N-1}x[i] is an efficient estimator of the mean.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
[2] Chatzichrisafis: “Solution of exercise 3.4 from Kay’s Modern Spectral Estimation - Theory and Applications”.