Given data

$$
\begin{aligned}
& I_{O O T}=5 \mu \mathrm{~A}, \\
& V_{C C}=30 \mathrm{~V}, \\
& R_1=30 \mathrm{k} \Omega
\end{aligned}
$$

By applying KVL to transistor $Q_1$ we can write

$$
\begin{aligned}
& -V_{C C}+I_{M N} R_1+V_{B E(\mathrm{ON})}=0 \\
& I_{M V}=\frac{V_{C C}-V_{B E(\mathrm{ON})}}{R_1} \\
& I_{M V}=\frac{30-0.7}{30 \times 10^3} \\
& I_{M V}=976.67 \times 10^{-6} \mathrm{~A}
\end{aligned}
$$


By applying KVL to base-emitter loop of both the transistors we can write

$$
\begin{aligned}
& -V_{B I 1}+V_{B E 2}+I_{O O T} R_2=0 \\
& V_{B E 1}=V_{B P 2}+I_{O U T} R_2 \\
& I_{O U T} R_2=V_{B E 1}-V_{B P 2}
\end{aligned}
$$


We have $V_{B B 1}-V_{B 22}=V_T \ln \left(\frac{I_{C 1}}{I_{O U T}}\right)$
Therefore $I_{O U T} R_2=V_T \ln \left(\frac{I_{C 1}}{I_{O U T}}\right)$

$$
\begin{aligned}
& R_2=\frac{V_T}{I_{O V T}} \ln \left(\frac{I_{C 1}}{I_{O V T}}\right) \\
& R_2=\frac{26 \times 10^{-6}}{5 \times 10^{-6}} \ln \left(\frac{976.67 \times 10^{-6}}{5 \times 10^{-6}}\right) \\
& R_2=27.428 \Omega
$ R_2=27.43 \Omega
$$


Output resistance is

$$
\begin{aligned}
& R_O=r_{o 2}\left(1+\frac{g_{m 2} R_2}{1+\frac{g_{m 2} R_2}{\beta_o}}\right) \\
& r_{o 2}=\frac{V_A}{I_D} \\
& r_{o 2}=\frac{130}{5 \times 10^{-6}} \\
& r_{o 2}=26 \times 10^6 \\
& r_{o 2}=26 \mathrm{M} \Omega
\end{aligned}
$$


Transconductance

$$
\begin{aligned}
g_{m 2} & =\frac{I_D}{V_F} \\
g_{m 2} & =\frac{1}{26} \times \frac{5}{1000} \\
g_{m 2} & =1.923 \times 10^{-4} \mathrm{~A} / \mathrm{V}
\end{aligned}
$$


The parameter $\beta_0$ we have is $\beta_0=200$
Substituting this values in $R_o$ we get

$$
\begin{aligned}
& R_O=26 \times 10^6\left(1+\frac{1.923 \times 10^{-4} \times 27.43 \times 10^3}{1+\frac{1.923 \times 10^{-4} \times 27.43 \times 10^3}{200}}\right) \\
& R_O=26 \times 10^6(6.134) \\
& R_O=159.499 \times 10^6 \Omega \\
& R_O=159.5 \mathrm{M} \Omega
\end{aligned}
$$