= Solution
Set
$$
F_\tau(x)=\sum_{\substack{\theta\in\Theta(R)\\\theta\subset\tau}}f_\theta(x)w_{r^{2/3}}(x).
$$
Freeze $x_3$. In the $(x_1,x_2)$ Fourier variables, the functions $F_\tau(\cdot,\cdot,x_3)$ are supported in caps of tangential length $r^{-1/3}$ and normal width $r^{-2/3}$ along the parabola. Apply part a with decoupling scale $R'=r^{2/3}$ and critical exponent $p=6$. Since $(R')^{1/2-3/6}=1$, for each fixed $x_3$,
$$
\left\|\sum_\tau F_\tau(\cdot,\cdot,x_3)\right\|_{L^6(\mathbb R^2)}^6
\lesssim_\epsilon r^\epsilon
\left(\sum_\tau\|F_\tau(\cdot,\cdot,x_3)\|_{L^6(\mathbb R^2)}^2\right)^3.
$$
Integrate in $x_3$. Minkowski's inequality in $L^3(dx_3)$ gives
$$
\int_{\mathbb R^3}\left|\sum_\theta f_\theta w_{r^{2/3}}\right|^6
\lesssim_\epsilon r^\epsilon
\left(\sum_{\tau\in\Theta(r)}
\left(\int_{\mathbb R^3}|F_\tau|^6\right)^{2/6}\right)^{6/2},
$$
which is exactly the claimed inequality after absorbing a change in $\epsilon$.
Solved by gpt-5.6-sol high.
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