$\alpha$ - particle

By angle chasing, we find that the required angle $\theta$ can be expressed as

$\theta = \arcsin \left( \dfrac{d}{R} \right)$

where $R = \dfrac{m_{\alpha} v}{q_{\alpha} B} = \dfrac{\sqrt{2 m_{\alpha} K}}{q_{\alpha} B}$ is the radius of the circular path followed by the alpha particle inside the magnetic field.

Thus

$\theta = \arcsin \left( \dfrac{q_{\alpha} B d}{\sqrt{2 m_{\alpha} K}} \right)$

Substituting

$\begin{cases} q_{\alpha} = 2 \times 1.6 \times 10^{-19} \text{ C} \\ B = 0.1 \text{ T} \\ d = 0.1 \text{ m} \\ m_{\alpha} = 4 \times 1.67 \times 10^{-27} \text{ kg} \\ K = \left( 2 \times 10^4 \right) \times 1.6 \times 10^{-19} \text{ J} \end{cases}$

we get

$\theta = \arcsin(0.4894) \approx \arcsin(0.5) = \boxed{30^{\circ}}$