A particle having a mass of 0.5 g carries a charge of $2.5\times {10}^{-8} \mathrm{C}$. The particle is given an initial horizontal velocity of $6\times {10}^4 {\mathrm{ms}}^{-1}$. To keep the particle moving in a horizontal direction

In the absence of a magnetic field, the particle will experience gravitational force mg. As a result, the particle will not continue moving in the horizontal direction but will describe a parabolic path. So, a magnetic field must be present and its direction must be perpendicular to the direction of the velocity. The magnetic force experienced by the particle is given by $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})$.
The magnitude of the force is $\mathrm{F}=\mathrm{qvBsin}\mathrm{\theta }$. If the particle is to move in the horizontal direction, this force must balance the force of gravity, i.e., $\mathrm{mg}=\mathrm{qvBsin}\mathrm{\theta }$
The minimum value of B corresponds to $\sin \mathrm{\theta }=1\text{ or }\mathrm{\theta }={90}^{\circ }$. Thus, $\mathrm{mg}=\mathrm{qvB}$

or $\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{qv}}=\frac{0.5\times {10}^{-3}\times 9.8}{2.5\times {10}^{-8}\times 6\times {10}^4}=3.27\mathrm{T}$

Hence, the correct options are (1) and (3).