I have found the CR-bound for the paremter of the exponential distribution $f(x) = 1/\lambda \exp(-x/\lambda)$
where $Var(T)\geq \frac{\lambda^2}{n}$.
I know how to the find the minimal sufficient statistics by ratio of likelihood functions: $\frac{L(X,\theta)}{L(Y,\theta)}, X = [X_1,...,X_n]',X_i,Y_i \sim_{i.i.d} Exp(\lambda)$, but i don't understands what it means and the significance to get a statistic with variance equal to the CR bound.
Let's consider the following statistics:
$$T=\overline{X}_n$$
It is well known and very easy to prove that
$$\mathbb{E}[T]=\lambda$$
Thus $T$ is unibased.
Now you can calculate variance for the sample mean and compare it with the lowest bound you stated.
$$\mathbb{V}[T]=\frac{\lambda^2}{n}$$
I edited you question because it is $\mathbb{V}[T]\geq \frac{\lambda^2}{n}$ where $T$ is an unbiased estimator and not $\mathbb{V}[\lambda]$ as you stated
Edit answering to your comments
$$\mathbb{E}[T]=\frac{1}{n}\mathbb{E}[\Sigma_i X_i]=\frac{1}{n}\cdot n \lambda=\lambda$$
If you do not want to do all the calculation there is also a Necessary and Sufficient condition which grants that the CR inequality is valid with $=$
$$\Sigma_i \frac{\partial}{\partial \lambda}\log f(x_i,\lambda)=b(n,\lambda)[T-g(\lambda)]$$
If you can factorize in this way, you can substitute $\geq$ with $=$ in CR inequality... (and of course you can, in your case)