What is the highest relative abundance for diatomic bromine?
In $\ce{Br2}$, the possible combinations of isotopes are:


$\ce{^{79}Br-^{79}Br} : \mathrm{m/z}~158$
$\ce{^{79}Br-^{81}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{79}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{81}Br} : \mathrm{m/z}~162$


Bromine's isotopic distribution is essentially a $50:50$ ratio of mass $79$ and mass $81$. That means that there's a $25\%$ chance of $\mathrm{m/z}~158$ $(0.5 \cdot 0.5 = 0.25)$ arising from $\ce{^{79}Br-^{79}Br}$. There's a $50\%$ chance of $\mathrm{m/z}~160$ since there are two combinations ($\ce{^{79}Br-^{81}Br}$ and $\ce{^{81}Br-^{79}Br}$) that give that mass $(0.5\cdot0.5 + 0.5\cdot0.5 = 0.25)$. There's a $25\%$ chance of $\mathrm{m/z}~162$  $(0.5\cdot 0.5 = 0.25)$ arising from $\ce{^{81}Br-^{81}Br}$.