Let's say the wire is in the $\overrightarrow x$ direction, while the applied magnetic field is in the $\overrightarrow z$ direction.
The Lorentz force law $\overrightarrow F=I\overrightarrow v\times \overrightarrow B $ tells us that the the force will be in the $\overrightarrow y$ direction.
The electrons in the wire will incur a net force in the direction perpendicular to their motion and the applied magnetic field. So if you do further calculations, you will see that the electrons will slow down a bit. If your wire is now a conducting plane, you'll have also a voltage difference across the conductor in the same direction of the force, see Hall effect
The resistance of the wire is an intrinsic parameter of the material of which it's made, so it cannot be modified by the magnetic field. As for the temperature, I think you'll need a strong alternating magnetic field for a non-negligible increase.
Below I'll use Planck units, for which, in particular, $c = \epsilon_{0} = =1$.
In fact, the full system of Maxwell's equations provides the statement that the only two vector components of the EM field $\mathbf E, \mathbf B$ are independent (in general, due to a deep symmetry reason, namely that a massless particle has only two polarizations). Next, if we write EM field in terms of 4-potential
$$
A_{\mu} \equiv \left( V, \mathbf A\right)_{\mu}
$$
(for example, $A_{0} \equiv V$),
$$
\mathbf E = -\frac{\partial \mathbf A}{\partial t} - \nabla V, \quad \mathbf B = \nabla \times \mathbf A,
$$
we add new unphysical gauge symmetry transformation
$$
\tag 1 V \to V + \frac{\partial \varphi}{\partial t}, \quad \mathbf A \to \mathbf A - \nabla \varphi ,
$$
under which $\mathbf E, \mathbf B$ is unchanged:
$$
\mathbf E \to \mathbf E, \quad \mathbf B \to \mathbf B
$$
I.e., $A_{\mu}$ is not unique. Also, it seems that without imposing some conditions 4-potential has 4 components, while this number must be 2, in correspondence to the number of independent components of $\mathbf E, \mathbf B$. Is this a problem? No.
We need to constuct the precise scheme of reduction of number of components, which includes the problem of uniqueness of 4-potential. The gauge symmetry is unphysical, so first we may fix $\varphi$ in $(1)$ by imposing so-called gauge condition, which leaves three independent components of $A_{\mu}$ instead four. It may be, for example, Coulomb gauge condition,
$$
\nabla \cdot \mathbf A = 0,
$$
or lorentz invariant condition, such as
$$
\partial_{\mu}A^{\mu} = \frac{\partial V}{\partial t} + \nabla \cdot \mathbf A = 0
$$
Next, we reduce the number of components from three to two by using first Maxwell equation, which is called Gauss law, since in fact it is the bound, because it doesn't contain first time derivatives of $\mathbf E$ or, correspondingly, second time derivatives of $A_{\mu}$ (it doesn't describe propagated degrees of freedom). For example, for Coulomb gauge this equation takes the form
$$
\nabla \cdot \mathbf E = -\Delta V - \frac{\partial (\nabla \cdot \mathbf A )}{\partial t} = -\Delta V = 4 \pi \rho ,
$$
from which we have the scalar potential as function of charge density. For $\rho = 0$, which is true in the case of magnetostatics, we may simply set $V$ to zero.
So that, we reduce the number of $A_{\mu}$ from four to two, as it must be, and solve the uniqueness problem.
Let summarize: if we have used the Gauss law as the bound, i.e., if we have reduced the number of components of 4-potential from four to three, the only thing that we need to fix it uniquely is to impose such gauge condition, since without doing that 4-potential is determined only up to unphysical transformation.
Best Answer
The question is probably simply asking you to write down the Lorentz force law, rather than rearrange it for $\mathbf{E}$ and $\mathbf{B}$ respectively. You could say: the magnetic flux density is the vector field $\mathbf{B}$ such that the force on a current $\mathbf{I}$ due to it is given by $\mathbf{F} = \mathbf{I} \times \mathbf{B}$.
The cross product, as you say, cannot be inverted. To see this, we note that the direction of $\mathbf{F}$ only tells us that $\mathbf{B}$ must lie in the plane perpendicular to $\mathbf{F}$. Then by the formula:
$$ |\mathbf{F}| = |\mathbf{I}||\mathbf{B}| \sin \theta \,, $$
we see that that the magnitude of $\mathbf{F}$ only pins down the value of $|\mathbf{B}|\sin \theta$, which involves two undetermined quantities. Hence we cannot determine $\mathbf{B}$. If we could invert the cross product, we would do something like this: consider that the cross product is linear, that is:
$$\mathbf{I} \times ( \alpha \mathbf{B}_1 + \beta \mathbf{B}_2) = \alpha \mathbf{I} \times \mathbf{B}_1 + \beta \mathbf{I} \times \mathbf{B}_2\,.$$
This means that we can write our cross product as a matrix equation:
$$ \mathbf{F} = \mathbf{I} \times \mathbf{B} \equiv \mathsf{M} \mathbf{B}\,. $$
Now what is the form of this matrix? To work this out, let's use suffix notation:
$$ F_i = \epsilon_{ijk} I_j B_k \,.$$
So we just have that
$$M_{ik} = \epsilon_{ijk} I_j\,. $$
At this point, you can check that the matrix $\mathsf{M}$ is not invertible, and so we cannot invert to give:
$$ \mathbf{B} = \mathsf{M}^{-1} \mathbf{F} \,,$$
as we would like. Consequently there's simply no way of writing $\mathbf{B}$ in terms of $\mathbf{I}$ and $\mathbf{F}$