= Solution
For the charge-generated <electric field>, use the <electric potential>
$$
\phi(\mathbf r,t)=\frac1{4\pi\varepsilon_0}\int\frac{\rho(\mathbf r',t)}{|\mathbf r-\mathbf r'|}\,dV',\qquad\mathbf E=-\nabla\phi.
$$
The <vector> potential is time independent, so its usual additional contribution $-\partial_t\mathbf A$ is zero. The <continuity equation> now gives
$$
\partial_t\mathbf E=-\frac1{4\pi\varepsilon_0}\nabla\int\frac{\dot\rho(\mathbf r',t)}s\,dV'=\frac1{4\pi\varepsilon_0}\nabla\int\frac{\nabla'\cdot\mathbf J(\mathbf r')}s\,dV'.
$$
Thus the second term in part (iii) is exactly the <displacement current> contribution $\mu_0\varepsilon_0\partial_t\mathbf E$, and
$$
\boxed{\nabla\times\mathbf B=\mu_0\mathbf J+\mu_0\varepsilon_0\partial_t\mathbf E.}
$$
This is the <Ampère-Maxwell equation>. The constructed field also has zero curl, consistent with the time-independent <magnetic field> and <Faraday's law>. A static addition to the <electric field> does not change this conclusion; the integral construction selects the fields sourced by the given densities rather than an independently added radiation field.
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