= Solution
Let $X^\mu(\tau)$ be the worldline, $u^\mu=dX^\mu/d\tau$ its <four-velocity>, and
$$
\Omega=\frac{qB_0}{m}.
$$
The <relativistic Lorentz force> in the constant background is
$$
\frac{du^\mu}{d\tau}
=\frac qm F^\mu{}_\nu u^\nu.
$$
Because $F^3=0$, its exponential terminates:
$$
u(\tau)
=\exp\left(\frac qmF\tau\right)u(0)
=\left[
I+\frac qmF\tau
+\frac12\left(\frac qmF\tau\right)^2
\right]u(0).
$$
The particle starts from rest, so
$$
u(0)=(c,0,0,0).
$$
Using the matrix from part (c) gives
$$
u^0=c\left(1+\frac12\Omega^2\tau^2\right),
\qquad
u^1=c\Omega\tau,
\qquad
u^2=-\frac c2\Omega^2\tau^2,
\qquad
u^3=0.
$$
Integrating from the origin and using $X^0=ct$ yields the parametric trajectory
$$
\boxed{
\begin{aligned}
t(\tau)&=\tau+\frac{\Omega^2\tau^3}{6},\\
x(\tau)&=\frac{c\Omega\tau^2}{2},\\
y(\tau)&=-\frac{c\Omega^2\tau^3}{6},\\
z(\tau)&=0.
\end{aligned}}
$$
This is the <relativistic trajectory in a constant null crossed field>. The real cubic relation for $t(\tau)$ is strictly increasing, so it determines $\tau$ uniquely as a function of coordinate time if an explicit $\mathbf x(t)$ is desired.
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