= Solution
Write the transverse displacement as $\mathbf s$ and take the common line-of-sight interval to be $[0,L]$. The <rotation measure> is a line integral of the <magnetic field>, so its <correlation> is
$$
C_{\mathrm{RM}}(\mathbf s)=a_0^2n_e^2\int_0^L dz\int_0^L dz'\,C_{zz}(\mathbf s,z-z').
$$
Substitute the <Fourier inversion> of the <solenoidal isotropic spectral tensor>. Keeping the finite observation length gives the exact windowed expression
$$
C_{\mathrm{RM}}(\mathbf s)=a_0^2n_e^2\int\frac{d^3k}{(2\pi)^3}H(k)\left(1-\frac{k_z^2}{k^2}\right)e^{i\mathbf k_\perp\cdot\mathbf s}\left|\int_0^L e^{ik_z z}\,dz\right|^2.
$$
The allowed extension of the longitudinal displacement integral to the whole real line is the <long-path projection of a magnetic correlation>. It applies when $L$ is much larger than the <correlation length>: equivalently, the squared window becomes $2\pi L\delta(k_z)$ inside the <integral>. The <Dirac delta function> sets $k_z=0$, leaving
$$
\boxed{C_{\mathrm{RM}}(s)\simeq a_0^2n_e^2L\int\frac{d^2k_\perp}{(2\pi)^2}H(k_\perp)e^{i\mathbf k_\perp\cdot\mathbf s}.}
$$
<Statistical isotropy> makes this a function of $s=|\mathbf s|$. The displayed two-dimensional relation is the intended long-path approximation; the preceding windowed formula accounts for finite-length edge effects.
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