= Solution
For $B=B(\Gamma,\rho)$ write $B_{1+\kappa}=B(\Gamma,(1+\kappa)\rho)$. It is a <Regular Bohr set> when
$$
|B_{1+\kappa}|=(1+O(d|\kappa|))|B|
$$
whenever $d=|\Gamma|$ and $|\kappa|\leq c/d$, with absolute constants in the $O$-term and in $c$.
Not every width is regular. In $G=\mathbb Z/3\mathbb Z$, let $\gamma(x)=e^{2\pi ix/3}$ and take $\rho=\sqrt3$. Then $B(\{\gamma\},\rho)=G$, but every arbitrarily small decrease of the width leaves only $0$. The size jumps from three to one, contradicting the required linear control as $\kappa\to0^-$.
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