Answer
$\bf Smaller.$
Work Step by Step
We need to find the radius of curvature of both rays so we can answer which has a greater deflection.
We know that the radius of curvature for a charged particle that is exposed to a magnetic field is given by
$$r=\dfrac{mv}{qB}$$
So,
$$\dfrac{r_{\alpha}}{r_{cathode}}=\dfrac{\dfrac{m_{\alpha}v_{\alpha}}{q_{\alpha}B}}{\dfrac{m_{cathode}v_{cathode}}{q_{cathode}B}}$$
where the two have the same velocity and are exposed to the same magnetic field.
$$\dfrac{r_{\alpha}}{r_{cathode}}= \dfrac{m_{\alpha} }{q_{\alpha} } \dfrac{q_{e} }{m_{e} } $$
where $q_\alpha=+2e$, $q_e=-e$, $m_e=9.11\times 10^{-31}$ kg, $m_\alpha=4 \times 1.67\times 10^{-27}$ kg
$$\dfrac{r_{\alpha}}{r_{cathode}}= \dfrac{4 \times 1.67\times 10^{-27} }{2e } \dfrac{e }{ 9.11\times 10^{-31}} =\bf 3666$$
$$r_\alpha\approx 3700\;r_{cathode}$$
The greater the radius of curvature, the smaller the deflection.
This means that the deflection of an alpha particle by a magnetic field is smaller than the deflection of a cathode-ray particle.
