The geometric distribution has density function $$f(x;\theta)=\theta(1-\theta)^x,$$
By definition, $I(\theta) = \operatorname{Var}(U)$ where $$U(x;\theta)= \frac{n}{\theta} - \frac{n\bar{x}}{1-\theta},$$ is the score statistic.
But then $$I(\theta)=\operatorname{Var}(\frac{n}{\theta} - \frac{n\bar{x}}{1-\theta})=(\frac{-n}{1-\theta})^2\frac{1}{n}\operatorname{Var}(x)= \frac{n}{(1-\theta)\theta},$$ which is apparently wrong, but I couldn't figure out where.
The steps to working out the Fisher information are as follows
Work out the likelihood. $$L=\prod_{i=1}^n\theta(1-\theta)^{x}=\theta^n(1-\theta)^{\sum_{i=1}^nx_i}.$$
Work out the log-likelihood. $$l=\log(L)=n\log(\theta)+\sum_{i=1}^nx_i\log(1-\theta).$$
Take two partial derivatives with respect to $\theta$. $$\frac{\partial l}{\partial\theta}=\frac{n}{\theta}-\frac{\sum_{i=1}^nx_i}{1-\theta},$$ then $$\frac{\partial^2 l}{\partial\theta^2}=-\frac{n}{\theta^2}-\frac{\sum_{i=1}^nx_i}{(1-\theta)^2}.$$
Finally, multiply through by $-1$ and take the expectation of this expression. $$F=\frac{n}{\theta^2}+\frac{n/\theta}{(1-\theta)^2}.$$
I'm sure you can simplify this accordingly.