= Solution
Use $x^0=t$ and the <Levi-Civita symbol> $\epsilon^1{}_{23}=1$. Define a smooth <affine connection> by
$$
\boxed{\Gamma^i{}_{00}=-E^i,\qquad \Gamma^i{}_{0j}=\Gamma^i{}_{j0}=\epsilon^i{}_{jk}B^k,\qquad \Gamma^0{}_{ab}=\Gamma^i{}_{jk}=0.}
$$
The lower indices are symmetric, so it is a <torsion-free connection>. With $t$ itself as parameter, the time component of the <geodesic equation> is identically satisfied, while its spatial components become
$$
\ddot x^i=-\Gamma^i{}_{00}-2\Gamma^i{}_{0j}\dot x^j=E^i+2\epsilon^i{}_{kj}B^k\dot x^j.
$$
The last term is $2(\mathbf B\times\dot{\mathbf x})^i$, with the wedge in the source interpreted as the three-dimensional <cross product>. Thus the spacetime lifts $(t,\mathbf x(t))$ are actually affinely parametrized <geodesics>, which proves the requested unparametrized claim. Conversely a geodesic with nonzero constant $dt/ds$ can be parametrized by $t$ and obeys this particle equation; purely spatial geodesics with $dt/ds=0$ are not such trajectories. No field equations or metric compatibility are needed. This construction is the <affine connection for a velocity-linear force>.
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