I don't understand exactly the problem given by your teacher.
The electromagnetic field is given by \vec{E} and \vec{B}, and these are the physically fields which give the force acting on charges according to Lorentz's Law, and that's how they are operationally defined (in the limit of negligible test charges).
For the fully time-dependent electromagnetic field you need a scalar and a vector potential. These are introduced due to the homogeneous Maxwell equations, i.e.,
\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0.
From these equations you draw the conclusion that there are a scalar field \Phi and a vector field \vec{A} such that
\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.
For a given electromagnetic field (\vec{E},\vec{B}), the potentials are not uniquely defined since obviously if (\Phi,\vec{A}) are a set of potentials that give the em. field, also
\Phi'=\Phi+\frac{1}{c} \partial_{\chi}, \quad \vec{A}'=\vec{A} - \nabla \chi
give the same em. field for an arbitrary scalar field \chi. Thus you can give one scalar constraint to simplify your task to solve for \Phi and \vec{A} from the remaining inhomogeneous Maxwell equations,
\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\vec{j}.
The first inhomogeneous equation (Gauß's Law), written in terms of the potential reads
-\left (\frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{A} + \Delta \Phi \right )=\rho.
If you choose the Lorenz-gauge condition
\frac{1}{c} \partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0,
then you have
\left (\frac{1}{c^2} \partial_t^2 - \Delta \right ) \Phi:=\Box \Phi=\rho.
For the Ampere-Maxwell Law you get in terms of the potentials
\vec{\nabla} \times (\vec{\nabla} \times \vec{A}) + \frac{1}{c} \left (\frac{1}{c} \partial_t^2 \vec{A}+\nabla \partial_t \Phi \right)=\vec{j}.
If you use the Lorenz-gauge condition again in
\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=-\frac{1}{c} \nabla \partial_t \Phi - \Delta \vec{A},
you get
\Box \vec{A}=\vec{j}.
As you see, the Lorenz gauge has the advantage that the equations for the scalar potential and the three components of the vector field decouple to four wave equations, and that's why this is a convenient gauge to describe the radiation from given charge and current distributions.
You can of course use any other gauge to solve for the inhomogeneous equations. For some tasks other gauges can be more convenient. The physical outcome, i.e., the em. field is of course always the same, independent of the gauge condition chosen.
