Here, $U = {A \over {{r^2}}} - {B \over r}$

For equilibrium, ${{dU} \over {dr}} = 0$

$\therefore - {{2A} \over {{r^3}}} + {B \over {{r^2}}} = 0$

$\Rightarrow {{2A} \over {{r^3}}} = {B \over {{r^2}}} \Rightarrow r = {{2A} \over B}$

For stable equilibrium, ${{{d^2}U} \over {d{r^2}}} \gt 0$

${{{d^2}U} \over {d{r^2}}} = {{6A} \over {{r^4}}} - {{2B} \over {{r^3}}}$

${\left. {{{{d^2}U} \over {d{r^2}}}} \right|_{r = \left( {2A/B} \right)}} = {{6A{B^4}} \over {16{A^4}}} - {{2{B^4}} \over {8{A^3}}} = {{{B^4}} \over {8{A^3}}} \gt 0$

So for stable equilibrium, the distance of the particle is ${{2A} \over B}$.