Given, two charge particles having same
charge, $q$ move with same speed i.e., $v_1=v_2=150 \mathrm{~km} / \mathrm{s}=1.5 \times 10^5 \mathrm{~m} / \mathrm{s}$
Electric force between two moving charge particle is same as they are static,
i.e.,
$\left|\mathbf{F}_2\right|=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}$
Magnetic force between two moving charge particles is given by
$\begin{aligned} \mathbf{F}_1 & =\frac{\mu_0}{4 \pi} \cdot \frac{q_1 \cdot q_2 v_1 v_2}{r^2} \\ {\left[\because q_1=q_2=q \text { and } v_1=v_2=v\right] } & \ldots \text { (ii) } \\ \left|\mathbf{F}_1\right| & =\frac{\mu_0}{4 \pi} \cdot \frac{q^2 v^2}{r^2}\end{aligned}$
From Eq. (i) and (ii), we get
$\begin{aligned} \therefore \quad \frac{\left|\mathbf{F}_1\right|}{\left|\mathbf{F}_2\right|} & =\frac{\frac{\mu_0}{4 \pi} \cdot \frac{q^2 v^2}{r^2}}{\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}}=\varepsilon_0 \mu_0 v^2 \\ & =\frac{1}{9 \times 10^{16}} \times\left(1.5 \times 10^5\right)^2=2.5 \times 10^{-7}\end{aligned}$
Hence, $\frac{\mathbf{F}_1}{\mathbf{F}_2}$ is $2.5 \times 10^{-7}$.