Because $V_{D C}=0, V_D=V_G$ and the FET is opcrating in the saturation region. Thus,

$$
\begin{aligned}
I_D & =\frac{1}{2} \mu_n C_{o x} \frac{W}{L}\left(V_{G S}-V_t\right)^2 \\
& =\frac{1}{2} \mu_n C_{o x} \frac{W}{L} V_{O V}^2
\end{aligned}
$$

from which we obtain $V_{O V}$ as

$$
\begin{aligned}
V_{o V} & =\sqrt{\frac{2 I_D}{\mu_n C_{o x}(W / L)}} \\
& =\sqrt{\frac{2 \times 80}{200 \times(4 / 0.8)}}=0.4 \mathrm{~V}
\end{aligned}
$$


Thus,

$$
V_{G S}=V_t+V_{O V}=0.6+0.4=1 \mathrm{~V}
$$

and the drain voltage will be

$$
V_D=V_G=+1 \mathrm{~V}
$$


The required value for $K$ can be found as follows:

$$
\begin{aligned}
R & =\frac{V_{D D}-V_D}{I_D} \\
& =\frac{3-1}{0.080}=25 \mathrm{k} \Omega
\end{aligned}
$$