Since the proton is entering the magnetic field at some angle other than 90^o, its path is helix.

Corresponding velocity of the proton along X-axis,

$v_x=v\cos 60°=2\times {10}^6\times \frac{1}{2}={10}^6{ \mathrm{ms}}^{-1}$

Due to velocity component v_x, the radius of the helix is described and is given by the relation

$r=\frac{m{v}_x}{qB}=\frac{1.6\times {10}^{-27}\times {10}^6}{1.6\times {10}^{-19}\times 0.10}=0.1 \mathrm{m}$

$\text{ Now, }T=\frac{2\pi r}{v_x}=\frac{2\pi \times 0.1}{{10}^6}=2\pi \times {10}^{-7} \mathrm{s}$