= Solution
For steady currents in vacuum, <Ampère's law> is $\oint\mathbf B\cdot d\mathbf l=\mu_0I_{\rm enclosed}$. A rectangular loop with one long side inside an ideal long <solenoid> and the other outside gives $B\ell=\mu_0(n\ell)I$, because the exterior field is negligible and the perpendicular sides contribute zero. Thus \b[$B=\mu_0nI$], approximately uniform away from the ends.
Write the <electron> <electric charge> as $q_e=-e$, with $e>0$. The <Lorentz force> is $\mathbf F=-e\mathbf v\times\mathbf B$; it does no work. For perpendicular motion, $mv^2/r_L=evB$ gives
$$
\boxed{\omega_c=\frac{eB}{m},\qquad r_L=\frac{mv}{eB}.}
$$
Here $\omega_c$ is the positive <angular speed>; the rotation sense is determined by the negative <electron> <electric charge>.
The parallel <velocity> is unaffected, while the perpendicular component rotates. Hence the path is a <helix> with $v_\parallel=v\cos\theta$ and $r_L=mv\sin\theta/(eB)$. For <on-axis entry into a solenoidal tube>, the transverse circle passes through that axis: its <guiding centre> is displaced by $r_L$, and its maximum distance from the tube axis is \b[$2r_L$], not $r_L$.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1-ia2012-electron-orbit.png]
{title=End-on electron orbit: the circle passes through the tube axis and reaches twice its Larmor radius from it}
Using the nonrelativistic approximation appropriate to $1\,\mathrm{keV}$,
$$
v=\sqrt{\frac{2E_0}{m}}=1.876\times10^7\,\mathrm{m\,s^{-1}},\qquad
\boxed{t=\frac{l}{v\cos3^\circ}=4.27\,\mathrm{ns}.}
$$
The <cyclotron period> is $2\pi/\omega_c=0.893\,\mathrm{ns}$, so the flight lasts $4.783$ revolutions. The <electron> therefore crosses the axis again \b[four times] before leaving. At $3^\circ$, $2r_L=0.279\,\mathrm{mm}<a$. Wall clearance requires $2mv\sin\theta/(eB)\le a$, giving
$$
\boxed{\theta_m=\sin^{-1}\!\left(\frac{eBa}{2mv}\right)=5.38^\circ.}
$$
The transit is longer than a full orbit, so the maximum transverse excursion really is sampled before exit.
For the <solenoidal time-of-flight energy window>, the prescribed time fixes $v_\parallel=l/t_0=v_0=\sqrt{2E_0/m}$, while $0\le v_\perp\le eBa/(2m)$. Therefore
$$
\boxed{E_{\min}=E_0=1.000\,\mathrm{keV},\qquad
E_{\max}=E_0+\frac{e^2B^2a^2}{8m}=1.0088\,\mathrm{keV}.}
$$
These are total <kinetic energies>. The upper value is the ideal grazing limit; if touching the wall counts as rejection, it is a <supremum> approached from below.
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