Calculating VDS in circuit
As you noted (devices in saturation), 

$$ I_{D1}=\dfrac{1}{2}k_{n1}(V_{GS1}-V_T)^2$$
Same goes for \$I_{D2}\$.

$$ I_{D2}=\dfrac{1}{2}k_{n2}(V_{GS2}-V_T)^2$$

And \$V_{GS1}=V_{IN}\$. They also gave you \$I_{D2}=I_{BIAS}=1\text{mA}\$. 

You could now use the equation for \$I_{D2}\$ and solve for \$V_{GS2}\$ (you have a value for everything else).

Keep in mind that \$V_{GS2}=V_X-V_{S2}\$. \$V_X\$ is easy to find since you have the current \$I_{D1}\$ and you have the value of \$R\$ too, it is indeed \$V_{DS1}\$. 

So once you know \$V_{GS2}\$ from the \$I_{D2}\$ equation, you can solve for \$V_{S2}\$ (knowing \$V_X\$). And finally \$V_{DS2}\$ is just the difference between \$V_{D2}\$ and \$V_{S2}\$.