I learned electrodynamics. According to the vector potential determination, $$ \mathbf B = [\nabla \times \mathbf A ], $$ Coulomb gauge, $$ \nabla \mathbf A = 0, $$ and one of Maxwell's equations, $$ [\nabla \times \mathbf B ] = \frac{1}{c}4\pi \mathbf j, $$
I can assume, that
$$ [\nabla \times \mathbf B ] = \nabla (\nabla \mathbf A) - \Delta \mathbf A = -\Delta \mathbf A = \frac{1}{c}4 \pi \mathbf j. $$ How to prove that the one of the solutions of this equation is solution like newtonian potential, $$ \mathbf A = \frac{1}{c}\int \limits_{V} \frac{\mathbf j (r) d^{3}\mathbf r}{|\mathbf r - \mathbf r_{0}|}? $$
Skipping the part about the Green's function, you should apply Fourier transformation on your equation
$$-\Delta \mathbf A = \frac{1}{c}4 \pi \mathbf j.$$
to change it into
$$k^2 \tilde{\mathbf{A}} = \frac{1}{c}4 \pi \tilde{\mathbf{j}}$$
or
$$\tilde{\mathbf{A}} = \frac{1}{c}4 \pi \tilde{\mathbf{j}}\cdot\frac{1}{k^2}$$
The left hand side is a product, so the inverse Fourier transform will be a convolution. The inverse Fourier transform of $1/k^2$ is $1/|r|$ in 3D, as you can see in formula 502. Therefore one gets
$$\mathbf A = \frac{1}{c}\int \limits_{V} \frac{\mathbf j (r) d^{3}\mathbf r}{|\mathbf r - \mathbf r_{0}|} \; .$$