A uniform magnetic field of magnitude 0.20 T is directed from east...

Let the directions be represented by unit vectors: East as $\hat{i}$, North as $\hat{j}$, and vertically upward as $\hat{k}$.

The magnetic field is directed from East to West, so $\vec{B} = -B \hat{i}$.

The particle is projected from South to North, so its velocity is $\vec{v} = v \hat{j}$.

The magnetic force on the particle is $\vec{F}_m = q(\vec{v} \times \vec{B}) = q(v \hat{j} \times (-B \hat{i})) = qvB \hat{k}$.

The magnetic force acts vertically upwards. For the particle to travel with uniform velocity, the net force must be zero. Thus, the upward magnetic force must balance the downward gravitational force:

$qvB = mg$

Substituting the given values $m = 0.010 \text{ g} = 1.0 \times 10^{-5} \text{ kg}$, $q = 1.0 \times 10^{-5} \text{ C}$, $B = 0.20 \text{ T}$, and $g = 9.8 \text{ m/s}^2$:

$v = \dfrac{mg}{qB} = \dfrac{1.0 \times 10^{-5} \times 9.8}{1.0 \times 10^{-5} \times 0.20}$

$v = \dfrac{9.8}{0.20} = 49 \text{ m/s}$