Cathode rays are composed of electrons, when they move in electric field a force

$F=eE$...(i)

Acts on them, this provides the necessary centripetal force to the particles

$F=\frac{m{v}^2}{r}$...(ii)

From Eqs. (i) and (ii), we get

$eE=\frac{m{v}^2}{r}$

$\Rightarrow r=\frac{m{v}^2}{eE}=\frac{m{\left({10}^6\right)}^2}{e\left(300\right)}$...(iii)

When velocity is doubled same circular path is followed, hence radius is same

$r=\frac{m{\left(2\times {10}^6\right)}^2}{eE}$...(iv)

Equating Eqs. (iii) and (iv), we get

$m\times \frac{{\left({10}^6\right)}^2}{300e}=\frac{m\times {\left(2\times {10}^6\right)}^2}{eE}$

$\Rightarrow E=300\times 4=1200$ V $\mathrm{c}{\mathrm{m}}^{-1}$