A particle with mass m and charge q is projected into a region...

The radius of the circular path of a charged particle in a perpendicular magnetic field is given by $R = \dfrac{mv}{qB}$.

When the particle enters the magnetic field perpendicularly to its boundary, the maximum horizontal distance it can penetrate into the region is equal to its radius $R$.

The given width of the magnetic field region is $d \approx \dfrac{2mv}{qB} = 2R$.

Since the width of the region $d$ is greater than the maximum penetration depth $R$ ($d > R$), the particle will not be able to cross the magnetic field region. Instead, it will complete a semicircular path inside the magnetic field and exit from the same face it entered.

Upon exiting, its velocity will be exactly opposite to its initial velocity (i.e., it turns back completely).

Therefore, the angle of deviation between the initial and final velocity vectors is $\pi$.