Assume $I_{REF} = I_O = 100 \mu A$ in both circuits, $\lambda = 0.01 V^{-1} $ for all transistors and $g_m = 0.5 mA/V$

Solution :-

The output resistance of the two-transistor current source is,

$r_o = \dfrac {1} {\lambda I_{REF}} = \dfrac {1} {(0.01)(0.10)} = 1M \Omega$

For the cascode circuit, we have $r_{o2} = r_{o4} = 1 M\Omega$. Therefore, the output resistance of the cascode circuit is,

$R_o = r_{o4} + r_{o2}(1 + g_{m}r_{o4})= 1+(1)[1+(0.5 \times 10^{-3})(10^6)] = 502M\Omega$

The output resistance of the cascode current source is substantially larger than that of the basic two-transistor circuit. Since $dI_o \alpha 1/R_o$, the load current in the cascode circuit is more stable against variations in output voltage.