$R_2=\frac{V_T(\ln 2)}{I_{\text {OUT }}}=\frac{(26 \mathrm{mV})(\ln 2)}{100 \mu \mathrm{~A}} \simeq 180 \Omega$
\begin{equation}
\begin{aligned}
&\begin{aligned}
\frac{1}{I_{\mathrm{OUT}}} \frac{\partial I_{\mathrm{OUT}}}{\partial T} & =\frac{1}{V_T} \frac{\partial V_T}{\partial T}-1500 \times 10^{-6}=\frac{1}{V_T} \frac{V_T}{T}-1500 \times 10^{-6} \\
& =\frac{1}{T}-1500 \times 10^{-6}
\end{aligned}\\
&\text { Assuming operation at room temperature, } T=300^{\circ} \mathrm{K} \text { and }\\
&\frac{1}{I_{\mathrm{OUT}}} \frac{\partial I_{\mathrm{OUT}}}{\partial T} \simeq 3300 \times 10^{-6}-1500 \times 10^{-6}=1800 \mathrm{ppm} /{ }^{\circ} \mathrm{C}
\end{aligned}
\end{equation}