Given, $E_y=900\sin \left(\omega t-\frac{\omega x}{c}\right)$

Here, $E_0=900$

Electric field and electric force are related as $F_E=q{E}_0$

Magnetic force of a moving charge is given by $F_B=qv{B}_0$

Thus, the ratio $\frac{F_E}{F_B}=\frac{E_0}{v{B}_0}$

Using relation between magnetic and electric field, we have

$\frac{F_E}{F_B}=\frac{c}{v}=\frac{3\times {10}^8}{3\times {10}^7}=10:1$