= Solution
Choose a <Regular Bohr set> $B'=B_\lambda\subseteq B$ with $1/2\leq\lambda\leq1$. Standard Bohr-set size estimates give $|B'|\geq2^{-O(d)}|B|$. Set $\eta=c_0/d$ with $c_0$ small enough that regularity gives
$$
|B'_{1-\eta}|\geq\frac12|B'|.
$$
For each $z\in B'_\eta$ and $y\in B'_{1-\eta}$, the triangle inequality in every frequency gives
$$
z-y,\ z, z+y\in B'.
$$
Because $|G|$ 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> gives
$$
|B'_\eta|\geq(\eta/8)^d|G|\geq(cd)^{-O(d)}|B'|.
$$
The number of progressions in $B$ is therefore at least
$$
|B'_\eta|\,|B'_{1-\eta}|
\geq(cd)^{-O(d)}|B'|^2
\geq(cd)^{-O(d)}|B|^2.
$$
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