In 2-bromo cyclohexanone,why bromine prefers to occupy axial rather than equatorial position?
I think there must have been some misunderstanding. This stabilization is working for all 2‑halocyclohexanones (but does not always cause the axial conformer to predominate), where bromo- one is not a special case, but has been used just as an example. 

There is a combination of various effects, but the dipolar one is supposed to cause a destabilization by dipolar repulsion in the equatorial-X conformer. See the difference of partial charges distribution:



Measured or calculated molar fractions (%) of the axial conformers in different phases/solvents (taken from Yoshinaga, Fabiana, et al. "Conformational analysis of 2-halocyclohexanones: an NMR, theoretical and solvation study." Journal of the Chemical Society, Perkin Transactions 2 9 (2002): 1494-1498. doi:10.1039/B204635K):

$$
\begin{array}{c*}\hline
&amp; \text{vapor} &amp; \text{liquid} &amp; \ce{CCl4} &amp; \ce{CHCl3} &amp; \ce{MeCN} &amp; \text{DMSO} \\
\hline
\ce{F}  &amp; 64 &amp;  5 &amp; 28 &amp; 13 &amp;  2 &amp;  2 \\
\ce{Cl} &amp; 86 &amp; 26 &amp; 62 &amp; 42 &amp; 15 &amp; 13 \\
\ce{Br} &amp; 92 &amp; 53 &amp; 81 &amp; 66 &amp; 36 &amp; 33 \\
\ce{I}  &amp; 96 &amp; 87 &amp; 92 &amp; 85 &amp; 65 &amp; 63\\
\end{array}
$$

Overall stabilization effects in axial-X conformation of 2‑halocyclohexanones clearly increases in the order of F &lt; Cl &lt; Br &lt; I.

However, the ‘gauche effect’ participates, which is stabilizing the equatorial conformation in largest extent for F, also for Cl, has almost zero effect for Br, and has a negative effect for I.