the angular displacement$\left(\theta \right),$
$\theta =\omega \text{t}$$\to \left(1\right)$

$\omega =$angular velocity, $t =$time taken by the the particle emerges on the other side 

$t = 10 \mathrm{ms} = 10\times {10}^{-3} \mathrm{S}$

angular velocity of particle in a magnetic field $\left(\omega \right)$,$\omega = \frac{qB}{m}$$\to \left(2\right)$

$q =$charge, $B =$magnetic field, $m =$mass of the particle.
substitute equation 2 in equation 1
$\theta =\frac{\text{qB}}{\text{m}}\text{t}$
Angle between velocity vectors is $\left(\theta \right)=\frac{\pi }{6}$
Solving above we get $\text{B}=\frac{50\pi \text{m}}{3\text{q}}$