Answer:

Step 1 - Calculate $I_{DQ}$

$V_G=\dfrac{R_2}{R_1+R_2} \times V_{DD}$

$V_G=\dfrac{110}{910+110} \times 20$

$\therefore V_G=2.156V$

Now , $V_S=I_{DQ}\times R_S=I_{DQ}\times 1.1$

$\therefore V_{GS}=(2.156-1.1I_{DQ})$

By schokley equation , 

$I_D=I_{DSS}[1-\dfrac{V_{GS}}{V_P}]^2$

$=10[1+\dfrac{2.156-1.1I_{D}}{3.5}]^2$

$1.21I_D^2-12.35I_D+25.60=0$

$\therefore I_D=7.31mA, \ I_D=2.89mA$

Now , $I_D\neq7.31mA$ because for this value of $I_D$ , $V_{DS}$comes out to be negative.

Hence , $ …