The potential on inner small sphere is $V_A$ = $V_{AA}+V_{AB}$ or $V_A$ = $\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]$ whereas the potential on the outer shell B is $V_B$ = $V_{BA}+V_{BB}$ = $\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]$ So, $V_A-V_B$ = $\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1}{r_1}-\frac{q_1}{r_2}\right]$ = $\frac{q_1}{4\pi\varepsilon_0}\left[\frac{r_2-r_1}{r_1r_2}\right]$ As $r_2$ > $r_1$, so $V_A$ > $V_B$ i.e. inner sphere A is at higher potential than outer conducting shell B, for any value of charge $q_1$. So, when inner sphere A is connected to outer shell B, then charge will flow from inner sphere A to outer shell B, until electric potentials on them is same i.e. $V_A-V_B$ = 0 or $q_1$ = 0 [As $r_1$ ≠ $r_2$] So, charge $q_1$ given to sphere A will flow on the shell B, no matter what the charge on the shell B is.