Equivalent resistance problem on cascode mirror circuit
At first glance, we can assume that:
$$V_{\pi3} \approx \frac{v_A}{2}$$
Do you know why we can do that?
So, we know the \$Q_1\$ emitter current value:
$$i_E \approx g_{m3}\frac{v_A}{2}$$
And \$Q1\$ base current will be equal to:
$$i_B = \frac{i_E}{\beta +1} = \frac{g_{m3}\frac{v_A}{2}}{\beta +1} = \frac{g_{m3} v_A}{2(\beta+1)} = v_A \frac{g_{m3}}{2(\beta+1)} $$
$$\frac{i_B}{v_A} = \frac{g_{m3}}{2(\beta+1)}$$
Therefore:
$$R_{eq} = \frac{v_{A}}{i_{B}} = \frac{v_A}{\frac{g_{m3} v_A}{2(\beta+1)}} = \frac{2 (\beta +1)}{g_{m3}}$$
And this is the end.