Find the Range of Rs for which the transistor is in saturation
For the device to be in saturation, we know that the drain-to-source voltage \$ V_{ds}\$ must satisfy \$ V_{ds} \geq V_{gs}-V_t \$, where \$V_{gs}\$ and \$V_t\$ are the gate-to-source voltage and threshold voltage of the device, respectively. For this particular problem, we can simplify the above inequality to be dependent on only the drain voltage \$ V_d\$ and \$ V_t\$ such that \$ V_d \geq -V_t \$, which implies that the drain current for the device \$I_d\$ satisfies 


  \$ I_d \leq \frac{5+V_t}{1 k\Omega}= I_{d,max} = 6 mA \$.


Let's assume that, for \$ R_s\ = 0\Omega\$, the device is in saturation. Under this assumption, you get a drain current of \$I_d = 1.6 A\$, which violates the above constraint on \$I_d\$. Thus we know that we seek to find a lower-bound on \$R_s\$. 

For a drain current of \$ I_d = I_{d,max} = 6 mA \$, you can make use of the \$I_d-V_{gs}\$ equation for a MOSFET in saturation to compute \$ V_{gs,max}\$ of the device to be \$1.245 V\$. By the use of KVL, we can obtain \$ V_{gs,max} + R_{s,min}I_{d,max} = 5\$. From there, you can compute the lower-bound on \$R_s\$ to be


  \$R_{s,min} = (5 - V_{gs,max})/I_{d,max} = 625.84 \approx 626 \Omega\$.


A concluding remark:

So long as \$R_s\$ satisfies \$ R_{s,min} \leq R_s &lt; \infty \$, the device will be in saturation; I encourage you to convince yourself that the device will also be on, i.e., \$ V_{gs} &gt; V_t \$. (In fact, for this particular circuit, \$ V_{gs} &gt; V_t \$ holds even if \$ 0 \leq R_s &lt; R_{s,min} \$.)