$$
\begin{gathered}
\text { Given } I_{\text {our }}=0.1 \mu \mathrm{~A} \\
R=10 \mathrm{k} \Omega
\end{gathered}
$$


From (3) we have $R=\frac{n V_T}{I_{I V}} \ln \left(\frac{I_{I N}}{I_{O V T}}\right)$

$$
\begin{aligned}
& 10 \times 10^3=\frac{1.5 \times 26 \times 10^{-3}}{I_{D V}} \ln \left(\frac{I_{D V}}{0.1 \times 10^{-6}}\right) \\
& 10 \times 10^3=\frac{0.039}{I_{D V}} \ln \left(\frac{I_{D V}}{0.1 \times 10^{-6}}\right) \\
& I_{D V}=3.9 \times 10^{-6} \times \ln \left(\frac{I_{D V}}{0.1 \times 10^{-6}}\right)
\end{aligned}
$$


The above equality is satisfied at $I_N=21 \mu \mathrm{~A}$
From (4) we have $\frac{W}{L}>\frac{I_{\mathrm{IN}}}{I_t}$

$$
\begin{aligned}
& \frac{W}{L}>\frac{21 \times 10^{-6}}{0.1 \times 10^{-6}} \\
& \frac{W}{L}>210
\end{aligned}
$$