= Solution
Since $L_z=J_\phi$, <Hamilton's equations> give
$$
\boxed{\dot L_z=-\frac{\partial H}{\partial\theta_\phi}=-\frac{\partial\delta\Phi}{\partial\theta_\phi}.}
$$
For the spherical unperturbed <Hamiltonian>, $H_0=H_0(J_r,L)$; in these particular <action-angle variables> the third component $\Omega_\phi=0$ describes the fixed orientation of the orbital plane. It should not be confused with the physical azimuthal <angular velocity> of a particle around its orbit.
The conserved <normalization of a probability distribution> $\int f\,d^3\theta\,d^3J=1$ makes the <expected value> of specific <angular momentum> equal to $\int L_zf\,d^3\theta\,d^3J$. Its time derivative is the <torque> per unit halo <mass>:
$$
\boxed{\mathcal T(t)=-\int f(\boldsymbol\theta,\mathbf J,t)\frac{\partial\delta\Phi}{\partial\theta_\phi}\,d^3\theta\,d^3J.}
$$
This follows either by averaging the single-particle equation or by integrating the <Collisionless Boltzmann equation> against $L_z$, with vanishing boundary fluxes.
The term $f_0\partial_{\theta_\phi}\delta\Phi$ integrates to zero over the periodic <angle variable>. Thus the first nonzero <torque> is quadratic in the perturbation. Use the <Fourier series> expansions of <3/b>; the source's reference here to part (a) is a numbering slip. Integrating the product over all three <angle variables> yields
$$
\begin{aligned}
\mathcal T(t)
&=-(2\pi)^3\sum_{\mathbf k,\boldsymbol\ell}i\ell_\phi
\int\delta f_{\mathbf k}\delta\Phi_{\boldsymbol\ell}\,
\delta_{\mathbf k+\boldsymbol\ell,\mathbf0}\,d^3J\\
&=\boxed{i(2\pi)^3\sum_{\mathbf k}k_\phi\int\delta f_{\mathbf k}(\mathbf J,t)\delta\Phi_{\mathbf k}^*(\mathbf J,t)\,d^3J.}
\end{aligned}
$$
Reality of the <potential energy> per unit <mass> gives $\delta\Phi_{-\mathbf k}=\delta\Phi_{\mathbf k}^*$; likewise $\delta f_{-\mathbf k}=\delta f_{\mathbf k}^*$. Conjugate harmonics make the total <torque> real. The sum is over all integer vectors, so no extra factor of two should be inserted.
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