Solution

Inside the magnetic field, the electron moves following a circular trajectory with constant speed.
From Newton Second Law:

$\frac{m_e v^2}{R} = e |\vec{v} \times \vec{B}|$

but $\vec{v}$ is perpendicular to $\vec{B}$, so

$\frac{m_e v^2}{R} = e v B$

$R = \frac{m_e v}{eB}$ (1)

Applying Energy Conservation Law:

$\frac{m_e v^2}{2} = e U$

$v = \sqrt{\frac{2eU}{m_e}}$ (2)

Substituting (2) into (1)

$R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$