= Solution
Write the <electron> <electric charge> as $-e$, with $e>0$ the <elementary charge>, and choose $\mathbf B=B\hat{\mathbf z}$ and initial <velocity> $u\hat{\mathbf x}$. In the nonrelativistic approximation, <Newton's second law> with the magnetic <Lorentz force> gives
$$
m_e\dot v_x=-eBv_y,\qquad m_e\dot v_y=eBv_x,\qquad \dot v_z=0.
$$
The magnetic <force> is perpendicular to the <velocity>, so it does no <work>: $d(m_e|\mathbf v|^2/2)/dt=0$. Setting $\omega_c=eB/m_e$, the solution with $x(0)=y(0)=0$ is
$$
v_x=u\cos\omega_ct,\qquad v_y=u\sin\omega_ct,
\qquad x=\frac u{\omega_c}\sin\omega_ct,\qquad
y=\frac u{\omega_c}(1-\cos\omega_ct).
$$
Consequently $x^2+(y-r)^2=r^2$, where $r=u/\omega_c$: this is a <cyclotron orbit>. Its <frequency> and <cyclotron radius> are
$$
\boxed{f=\frac{eB}{2\pi m_e},\qquad r=\frac{m_eu}{eB},\qquad
u=\sqrt{\frac{2eV}{m_e}}.}
$$
The accelerating <voltage> changes $u$ and $r$ but cancels out of $f$. Using $m_e=9.1094\times10^{-31}\,\mathrm{kg}$ and $e=1.6022\times10^{-19}\,\mathrm C$ gives
$$
B=\frac{2\pi m_e(2.5\times10^9\,\mathrm{s^{-1}})}e
\simeq0.0893\,\mathrm T,\qquad
u\simeq1.876\times10^7\,\mathrm{m\,s^{-1}},
\qquad\boxed{r\simeq1.19\times10^{-3}\,\mathrm m.}
$$
Only the magnitude of the <magnetic field> is determined; its direction is perpendicular to the orbital plane, with the sense of rotation fixed by the negative <electric charge>. The claimed voltage independence is a nonrelativistic result: exactly, $\gamma=1+eV/(m_ec^2)$ and $f=eB/(2\pi\gamma m_e)$. At $1\,\mathrm{kV}$ the correction is about $0.2\%$. The stated model also neglects the <energy> lost to emitted <electromagnetic radiation>.
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