Answer: Step 1 : Calculate the value of k :

$k=\dfrac{I_D(on)}{[V_{GS}(on)-V_{TH}]^2}$

$\therefore k=\dfrac{6\times 10^{-3}}{(8-3)^2}$

$\therefore k=0.24 \times 10^{-3}A/V^2$...................(1)

Step 2 : obtain $I_{DQ}$

$I_D=k[V_{GS}-V_{TH}]^2$..........................(2)

Now , $V_{GS}=V_{DD}-I_DR_D$

$V_{GS}=12-I_DR_D$......................(3)

Putting the value of (1) and (3) in (2),

But $R_D=2k=2000 \Omega$

$\therefore I_D=0.24 \times 10^{-3}[9-2000I_D]^2$

$\therefore I_D=0.24 \times 10^{-3}[81-36 \times 10^3I_D+4 \times 10^6I_D^2]$

$\therefore I_D=0.01944-8.64 \times I_D+960I_D^2 $

$\therefore 960I_D^2-9.64I_D+0.01944=0$

Solving the above quadratic equation ,we get 

$\therefore I_D=7.24mA \ or \ I_D=2.79mA$

Now $I_D\neq7.24mA$ because if $I_D=7.24mA$ the value of $V_{DS}$ becomes negative.

Hence , $I_{DQ}=2.79mA$

Step 3 : Calculate $V_{DSQ}$

$V_{DSQ}=V_{DD}-I_DR_D$

$\therefore V_{DSQ}=12-(2.79 \times 2)$

$\therefore V_{DSQ}=6.42 V$