In a tube like the one J. J. Thomson used, electrons go straight through crossed fields of 4.0 ×…

Step 1. The speed. The beam goes straight, so $eE = evB$. We want $v$, so $v = \dfrac{E}{B} = \dfrac{4.0 \times 10^{4}\ \text{V/m}}{4.0 \times 10^{-3}\ \text{T}} = 1.0 \times 10^{7}\ \text{m/s}$, that is $10.0 \times 10^{6}$ m/s.

Step 2. The ratio. With the electric field off, the circle has $r = \dfrac{mv}{eB}$. We want $e/m$, so $\dfrac{e}{m} = \dfrac{v}{Br}$. The bottom is $Br = 4.0 \times 10^{-3}\ \text{T} \times 1.42 \times 10^{-2}\ \text{m} = 5.68 \times 10^{-5}\ \text{T m}$. So $\dfrac{e}{m} = \dfrac{1.0 \times 10^{7}\ \text{m/s}}{5.68 \times 10^{-5}\ \text{T m}} = 1.76 \times 10^{11}\ \text{C/kg}$.