In a vacuum tube, an electron moves at 1.8 × 10⁷ m/s at right angles to a uniform field of 0.025 T…

Step 1. The force. $F = evB = 1.6 \times 10^{-19}\ \text{C} \times 1.8 \times 10^{7}\ \text{m/s} \times 0.025\ \text{T} = 7.2 \times 10^{-14}\ \text{N}$.

Step 2. The acceleration. We want $a$, so $a = \dfrac{F}{m} = \dfrac{7.2 \times 10^{-14}\ \text{N}}{9.1 \times 10^{-31}\ \text{kg}} = 7.91 \times 10^{16}\ \text{m/s}^2$.

Step 3. The weight. $mg = 9.1 \times 10^{-31}\ \text{kg} \times 9.8\ \text{m/s}^2 = 8.9 \times 10^{-30}\ \text{N}$. The ratio is $\dfrac{7.2 \times 10^{-14}\ \text{N}}{8.9 \times 10^{-30}\ \text{N}} = 8.07 \times 10^{15}$.