When the bar magnet is turned through 180°, neutral points would lie on equatorial line, so that $B_2$ = $\frac{\mu_0}{4\pi} \frac{M}{d_2^3}$ = $B_H$ ... (i) In the previous question, $B_1$ = $\frac{\mu_0}{4\pi} \frac{2M}{d_1^3}$ = $B_H$ ... (ii) From (i) and (ii) $\frac{\mu_0}{4\pi} \frac{M}{d_2^3}$ = $\frac{\mu_0}{4\pi} \frac{2M}{d_1^3}$ ∴ $d_2^3$ = $\frac{d_1^3}{2}$ = $\frac{(14)^3}{2}$ or $d_2$ = $\frac{14}{2^{1/3}}$ = 11.1 cm