= Solution
For each $\tau\in\Theta(r)$ put
$$
F_\tau=\sum_{\substack{\theta\in\Theta(R)\\\theta\subset\tau}}f_\theta w_{r^{1/3}}.
$$
Multiplication by $w_{r^{1/3}}$ enlarges Fourier support by at most $r^{-1/3}$ because $\widehat w_{r^{1/3}}$ is supported in $B_{r^{-1/3}}(0)$. The supports of the $F_\tau$ therefore have uniformly bounded overlap: separated $r^{-1/3}$-scale moment-curve intervals remain disjoint except for boundedly many neighbors.
The <fourier-support almost orthogonality> supplied by the Plancherel theorem now gives
$$
\int_{\mathbb R^3}\left|\sum_{\theta\in\Theta(R)}f_\theta w_{r^{1/3}}\right|^2
\lesssim
\sum_{\tau\in\Theta(r)}\int_{\mathbb R^3}
\left|\sum_{\substack{\theta\in\Theta(R)\\\theta\subset\tau}}f_\theta w_{r^{1/3}}\right|^2.
$$
Solved by gpt-5.6-sol high.
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