this I just want to add some details to noob2 comments and also answer your second question inside the comments.
So first lets consider a standard normal distribution $X_0$. This means that $X_0$ has mean zero ($\mu = 0$) and standard deviation 1 ($\sigma = 1$). Let's write $ \Phi^{-1}(\alpha)$ for the $\alpha$ quantile of $X_0$.
In R you can compute $ \Phi^ {-1}(\alpha)$ using qnorm
, i.g. $\Phi^{-1}(0.1)$ = qnorm(0.1)
.
Now consider $X \sim \mathcal{N}(\mu, \sigma)$. Then we have that
$$
X \stackrel{\text{d}}{=} \mu + \sigma \cdot X_0.
$$
As a consequence the quantiles of $X$ can be written as $\Phi^{-1}(\alpha) \cdot \sigma + \mu$. You can easily verify this in R:
mu <- 3
sigma <- 1.2
alpha <- 0.45
qnorm(0.45) * sigma + mu
[1] 2.849206
qnorm(0.45, mean = mu, sd = sigma)
[1] 2.849206
Therefore
$$
\frac{90th quantilie - 10th quantile}{\sigma} = \frac{\Phi^{-1}(0.9) \cdot \sigma + \mu - \Bigl(\Phi^{-1}(0.1) \cdot \sigma + \mu \Bigr)}{\sigma} =\\ \Phi^{-1}(0.9) - \Phi^{-1}(0.1).
$$
I hope this answers your question from the comments.