finding the output resistance of bjt folded cascode
The first thing you should notice is that from a small-signal preceptive point of view \$Q_1\$ \$r_{o1}\$ act just like the emitter resistance for \$Q_2\$

And the small signal mode will look like this 







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The small-signal parameters are: 

\$\large g_{m2} = \frac{0.1\textrm{mA}}{25\textrm{mV}} = 4\textrm{mS}\$ 

\$\large r_{\pi 2} = \frac{\beta}{g_{m2}} = 25\textrm{k}\Omega\$

\$\large r_{o2} = r_{o1} = \frac{V_A}{\textrm{I}} =\frac{5 \textrm{V}}{0.1\textrm{mA}} = 50 \$

Hence the output resistance will be the same as I show here:

BJT common-base output resistance derivation

finding output resistance of CB amplifier with ro

$$R_{OUT} = \frac{V_X}{I_X} = r_{o2}+\left(1+ g_{m2} r_{o2}\right)\cdot r_{o1}||r_{\pi2} = 3.4\textrm{M}\Omega $$

Also, remember that in the small-signal analysis you can replace PNP with the NPN small-signal model.  







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simulate this circuit