 Solution

Choose a Regular Bohr set $B^{\prime }=B_{\lambda }\subseteq B$ with $1/2\le \lambda \le 1$. Standard Bohr-set size estimates give $|B^{\prime }|\ge 2^{-O(d)}|B|$. Set $\eta =c_0/d$ with $c_0$ small enough that regularity givesFor each $z\in B_{\eta }^{\prime }$ and $y\in B_{1-\eta }^{\prime }$, the triangle inequality in every frequency givesBecause $|G|$ is odd, multiplication by two is a bijection, and the pair $(z,y)$ determines the ordered three-term arithmetic progression uniquely. The lower size bound for a Dilate of a Bohr set givesThe number of progressions in $B$ is therefore at least