= Solution
<Statistical homogeneity> makes the two-point <correlation> depend on the displacement $\mathbf u=\mathbf r-\mathbf r'$. Substituting the <Fourier transforms> and changing variables from $(\mathbf r,\mathbf r')$ to $(\mathbf u,\mathbf r')$ gives
$$
\begin{aligned}
\mathbb E[\widehat B_i(\mathbf k)\widehat B_j(\mathbf k')]
&=\int d^3u\,d^3r'\,C_{ij}(\mathbf u)e^{-i\mathbf k\cdot\mathbf u}e^{-i(\mathbf k+\mathbf k')\cdot\mathbf r'}\\
&=\widehat C_{ij}(\mathbf k)\int d^3r'\,e^{-i(\mathbf k+\mathbf k')\cdot\mathbf r'}.
\end{aligned}
$$
The last <integral> is a <Dirac delta function>, so
$$
\boxed{\mathbb E[\widehat B_i(\mathbf k)\widehat B_j(\mathbf k')]=(2\pi)^3\delta(\mathbf k+\mathbf k')\widehat C_{ij}(\mathbf k).}
$$
The plus sign occurs because neither <Fourier transform> has undergone <complex conjugation>. For a real <magnetic field>, the version with $\widehat B_j(\mathbf k')^*$ instead has $\delta(\mathbf k-\mathbf k')$. Homogeneity determines dependence on the displacement vector; dependence only on its <norm> needs <statistical isotropy> as well.
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