MOSFET: Calculating \$V_\mathit{DSsat}\$ from \$k_n'\frac{W}{L}\$ and \$V_\mathit{Th}\$
No, it is impossible to find \$V_{DS_{sat}}\$ based on the parameters you've provided.
Theory:
The most basic model for representing NMOS's current is this:

Due to the fact that both \$C_{i}\$ and \$\mu\$ are parameters of a particular technology and are constant across all the NMOSs in a given technology, it is common to replace two constants with a single one: \$k'=C_{i}\mu\$.
If you plot the above equation, you'll find something strange - it predicts a maximum of \$I_{D}(V_{DS})\$. It means that there is some \$V_{DS_{sat}}\$, and for \$V_{DS}&gt;V_{DS_{sat}}\$ the current is decreasing! There might be two explanations to this phenomenon:

There is some very unusual effect takes place.
The above equation has limited validity (in terms of \$V_{DS}\$).

The second bullet is the correct one - this equation is valid up to \$V_{DS}=V_{DS_{sat}}\$. When this threshold is reached, the conducting channel underneath transistor's Gate &quot;pinches-off&quot; and the current does not increase anymore with increasing \$V_{DS}\$:

The names of the regions of operation appear on the graph. Linear region is sometimes referred to as &quot;triode region&quot;.
So, how one finds an expression for \$V_{DS_{sat}}\$? Very simple: differentiate the above equation with respect to \$V_{DS}\$ and find when the derivative equals to zero. You'll get the following value:
$$V_{DS_{sat}} = V_{GS}-V_T&gt;0$$
The last inequality represents the fact that the transistor is not in cut-off region.
Substituting this value back to current's equation, you'll find:
$$I_{D_{sat}}=\frac{1}{2}k'\frac{W}{L}(V_{GS}-V_T)^2$$
Now, back to your question:
As you can see, substituting all the given parameters into the last equation allows you to calculate \$V_{GS}\$, and, since you also know \$V_T\$, you can calculate \$V_{DS_{sat}}\$. However, this requires one additional assumption which you did not state in the question: the transistor should be known to operate in saturation region.
Otherwise, if the transistor is in linear region, you need to know also at which \$V_{DS}\$ it operates in order to be able to calculate \$V_{GS}\$ from the first equation.