Step 1: Bring in the working rules for slow, quasi-static EM fields.
For EM induction studies of the Earth, the fields change slowly enough that we drop the displacement current term. This leaves four rules to check each statement against: charge conservation, no magnetic monopoles, the link between $\mathbf{B}$ and $\mathbf{A}$, and Faraday's law.
Step 2: Test the magnetic field claim first.
Magnetic field lines always close on themselves, they never start or stop at some point charge equivalent. This is written as $\nabla \cdot \mathbf{B} = 0$, which is exactly what "solenoidal" means. So the claim that $\mathbf{B}$ is solenoidal is right.
Step 3: Test the current density claim.
Inside a conductor with finite conductivity, under the slow field approximation used here, charge build up over time is negligible, so $\nabla \cdot \mathbf{J} = 0$ as well. This again fits the definition of solenoidal, so this claim is right too.
Step 4: Test the vector potential claim.
The vector potential $\mathbf{A}$ is defined through $\mathbf{B} = \nabla \times \mathbf{A}$. Saying $\mathbf{A}$ is irrotational would mean its curl is zero, but its curl is $\mathbf{B}$, which is not zero. So this claim is false.
Step 5: Test the curl of E claim.
Faraday's law says $\nabla \times \mathbf{E} = -\dfrac{\partial \mathbf{B}}{\partial t}$. Writing $\mathbf{B}$ as $\nabla \times \mathbf{A}$ turns this into $\nabla \times \mathbf{E} = -\dfrac{\partial}{\partial t}(\nabla \times \mathbf{A})$, which is exactly the statement given, so this claim is right too.
Step 6: Collect the correct ones.
\[ \boxed{\text{(A), (C) and (D)}} \]