 Solution

For $B=B(\Gamma ,\rho )$ write $B_{1+\kappa }=B(\Gamma ,(1+\kappa )\rho )$. It is a Regular Bohr set whenwhenever $d=|\Gamma |$ and $|\kappa |\le c/d$, 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 $\rho =\sqrt{3}$. 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 $\kappa \to 0^{-}$.