b
\(\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,Here,\,u = \frac{A}{{{r^2}}} - \frac{B}{r}\\
For\,equilibrium,\,\frac{{du}}{{dr}} = 0\\
\therefore \,\, - \frac{{2A}}{{{r^3}}} + \frac{B}{{{r^2}}} = 0\,or\,\frac{{2A}}{{{r^3}}} = \frac{B}{{{r^2}}}\,or\,r = \frac{{2A}}{B}\\
For\,stable\,equilibrium,\,\frac{{{d^2}u}}{{d{r^2}}} > 0
\end{array}\)

\(\begin{array}{l}
\frac{{{d^2}u}}{{d{r^2}}} = \frac{{6A}}{{{r^4}}} - \frac{{2B}}{{{r^3}}}\\
{\left. {\frac{{{d^2}u}}{{d{r^2}}}} \right|_{r = \left( {2A/B} \right)}}\, = \frac{{6A{B^4}}}{{16A}} - \frac{{2{B^4}}}{{8{A^3}}} = \frac{{{B^4}}}{{8{A^3}}} > 0\\
So\,for\,stable\,euilibrium,\,the\,{\rm{distance}}\\
{\rm{of}}\,{\rm{the}}\,{\rm{particle}}\,{\rm{is}}\,\frac{{2A}}{B}
\end{array}\)