Is the 110 m length accurate for 3.0 mm 1 kg spool?
Let's go about this scientific:

$A_r=\pi{r}^{2}=\pi{\frac{d}{2}}^{2}$

As a result the crossections are $A_{3}=7.06858\text{mm}^2=0.0707\text{cm}^2$, $A_{2.85}=6.3794\text{mm}^2=0.0638\text{cm}^2$ and $A_{1.75}=2.40528\text{mm}^2=0.024\text{cm}^2$.

Volume of a cylinder is $V_{A_d,l}=\times {A_d} {l}$. Turned around to get a length from Volume and Area we get $l=\frac {V_m}{A_d}$, but what is V?

We know the density of comercial PLA is about $\rho=1.25 \frac{\text g}{\text{cm}^3}$, and we know $m=\times V \rho$. So: $V_m=\frac{m}{\rho}=\frac{1000}{1.25}\text{cm}^3=800\text{cm}^3$.

Taking this Volume and using the $l=\frac {V_m}{A_d}$ we get:

$l_{d=1.75}=33333.33\frac{\text{cm}}{\text{kg}}=333.33\frac{\text{m}}{\text{kg}}$

$l_{d=2.85}=12539.18\frac{\text{cm}}{\text{kg}}=125.39\frac{\text{m}}{\text{kg}}$

$l_{d=3}=11315.41\frac{\text{cm}}{\text{kg}}=113.15\frac{\text{m}}{\text{kg}}$

If the filament is more on the dense side $(\rho&gt;1.25\frac{\text g}{\text{cm}^3})$, then it will have a smaller volume and thus be shorter than this estimate.

To show this better, a graph: This is the length of a filament spool in dependancy of the density. The values were calculated for the usual diameters with their closest neighbors rounded to 0.1 as absolute diameters and run over a broad range of densities commonly used in plastics - 0.7 g/cm³ to 2 g/cm³.