Absorption coefficient for liquid water
What the plot shows is the reciprocal path length needed for a given absorbance to apply, for example slightly less than 100 m in the visible to $10^{-8}$ m in the uv, as you point out, but this is for $55.345$ molar water which is  very, very concentrated compared to most solutions from which an absorption spectrum is obtained, typically $&lt;&lt; 1 $ molar.

As you also point out the absorbance is $A=\log_{10}(I_0/I)=\epsilon[C]l$ with $\epsilon$ as the extinction coefficient usually in units $\pu{dm^3mol^{-1}cm^{-1}}$ and which depends upon wavelength, and $[C]=55.345 \pu{ mol}$ for pure water and l the path length in cm. What seems to be plotted is 
$$\frac{1}{l}=\frac{\epsilon[C]}{A}$$
so is still proportional to the extinction coefficient and so still indicates regions where water absorbs greatly or otherwise. But it is not clear what value for A is used. It may be understood that $A=1$ always applies or that the value is taken at $1/e$ absorption. The values may also be based on $\log_e$ not $10$ so a factor of $2.303$ may be involved. As I'm not familiar with this much detail I can't help any more other than to suggest that  you will have to check by finding a true absorption spectrum or  extinction coefficient and work backwards and converting units as necessary. 

In general the amount of absorption a molecule has can vary a huge amount. In water in the visible part of the spectrum absorption is low as there are only weak overtone bands from vibrational transitions and nothing else. At shorter wavelengths electronic absorption begins and these can have extinction coefficients of tens to hundreds of thousands in units of $\pu{dm^3mol^{-1}cm^{-1}}$. 

A  concentrated solution of a dye  at, say, $1$ molar with a typical extinction coefficient of $10^4$ would have an optical density of $4$ in a $1$ cm cell or a reciprocal length of $10^6$ $\pu{m^{-1}}$ if $A=1$. Hope this helps.