Waves and guitar frets

The frequency spectrum is given by:

$$f_n=frac{n}{2L}sqrt{frac{T}{rho}}$$

and:

$$f_1=frac{1}{2L}sqrt{frac{T}{rho}}$$

where $T$ is the string tension.

Alternatively, write:

$$f_n=frac{1}{2x(n)}sqrt{frac{T}{rho}}$$ So that:

$$boxed{frac{f_n}{f_1}=frac{x(n)}{L}}$$

Where $x(1)=L$.

Calculate the $x(n)$ from there.

Take a look at the picture bellow (adapted from here https://guitar-auctions.co.uk/wp-content/uploads/2018/02/lot0126.jpg) where I give you the recursive idea to calculate the length $x_{n+1}$ from $x_n$ for the frets on a guitar. $x_0$ is simply the length of your string (in your case $x_0 = L = 1 $ m). The density of the string does not matter. Why? As you can see, the guitar has six strings. All string have different densities, but the fret distance progression is the same for all strings.

The density, together with the length $L$, has a role in the fundamental frequency of the string which is given by

$$f_1 = frac{1}{2L} sqrt{frac{T}{mu}}$$

As can be seen, the product of the fundamental frequency by the length is a constant, because $T$ (string tension) and $mu$ (string density) are constants. So by changing the length $x_0 = L$ of the string to $x_1$ will change the frequency. That is

$$ f_1 L = f_2 x_1 $$

But the frequency $f_2$ should be $f_2 = f_1 2^{1/12}$ according to the exercise you solved. So substituting in the above equation and eliminating $f1$ we find

$$ L = 2^{1/12} x_1$$

having into account that we defined $ L = x_0$ we find

$x_1 = frac {x_0}{2^{1/12}}$

You can proceed the same way to calculate $x_{n+1}$ from $x_n$.