a
As the initial velocity of the particle is perpendicular to the field, the particle will move along the arc of a circle as shown (fig). If $r$ is the radius of the circle, then
$\frac{m v_{0}^{2}}{r}=q v_{0} B_{0} \Rightarrow r=\frac{m v_{0}}{q B_{0}}$
Also, from geometry, $L=r \sin 30^{\circ} \Rightarrow r=2 L$
or $L=\frac{r}{2}=\frac{m v_{0}}{2 q B_{0}}$
