Calculating a NRR of an IIR
Let's find the impulse response for this system:

$$H(z)= \frac{z^{-5}}{1-0.1z^-1} = z^{-5}\frac{1}{1-0.1z^-1}$$

You didn't specify the ROC, so I'll assume that we take it so that the filter is stable. Namely:

$$|z|&gt;\frac{1}{10}$$

I've split the expression of $H(z)$ to notice that the impulse response basically consists of an exponential with a delay. We know that

$$\frac{1}{1-0.1z^-1}\xrightarrow{\mathcal{Z}^{-1}}(0.1)^nu[n]$$

where $u[n]$ is the unitary step. The $z^{-5}$ factor translates to a delay in the time domain. So the impulse response is:

$$h[n]=(0.1)^{n-5} u[n-5]$$

If we want to find the sum of all the terms squared in the sequence:

$$\begin{align}
\sum_{n=-\infty}^{\infty}h^2[n]
&amp;=\sum_{n=-\infty}^{\infty}\left[(0.1)^{n-5} u[n-5]\right]^2\\
&amp;=\sum_{n=5}^{\infty}\left[(0.1)^{n-5}\right]^2\\
&amp;=\sum_{m=0}^{\infty}(0.1)^{2m}\\
&amp;=\sum_{m=0}^{\infty}(0.01)^{m}\\
&amp;=\frac{1}{1-0.01}\\
&amp;=\frac{100}{99}\\
\end{align}$$

Where I've used the formula for the infinite geometric sum.

EDIT:

Note that a smaller NRR implies a worse performance of the filter at reducing noise. In this case, NRR turned out to be almost $1$ (or $0 \ \mathrm{dB}$), so the variance of the noise at the output is practically the same as in the input. As Fat32 pointed out in the comments, this depends on the position of the pole of your filter. The noise is white (as far as I understood from your comments), so it has infinite bandwidth. If the filter's bandwidth is too large, then "a lot" of noise will pass through it, appearing at the output. On the contrary, if it is narrow, then it will "filter out" most of it, leading to a decrease in the variance at the output. In this case, the filter's bandwidth is rather wide due to the pole being near the origin, and then the NRR approaches $1$: not a really good filter to reduce noise.