For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore,

$$\begin{gathered} cz\ge \frac{\sigma q}{2{ϵ}_0}\left[R-\left(\sqrt{R^2+z^2}-z\right)\right] \\ \Rightarrow \frac{c2{ϵ}_0}{\sigma q}\ge \frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z} \\ \Rightarrow \beta \ge \frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z} \end{gathered}$$

For option A

$\frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z}=\frac{R-R\left(\sqrt{1+{\displaystyle \frac{625}{49}}}-{\displaystyle \frac{25}{7}}\right)}{R{\displaystyle \frac{25}{7}}}\simeq 0.242$

Since, $0.25>0.242$

Option A is correct.

For option B

$\frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z}=\frac{R-R\left(\sqrt{1+{\displaystyle \frac{9}{49}}}-{\displaystyle \frac{3}{7}}\right)}{R{\displaystyle \frac{3}{7}}}\simeq 0.79$

Since, $0.25<0.79$

Option B is incorrect

For option C

$\frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z}=\frac{R-R\left(\sqrt{1+{\displaystyle \frac{1}{3}}}-{\displaystyle \frac{1}{\sqrt{3}}}\right)}{R{\displaystyle \frac{1}{\sqrt{3}}}}\simeq 0.73$

Since, $0.25<0.73$

Therefore, the particle returns to $z_0$

Hence, option C is correct.

For option D,

For any value of $z>0$

$\frac{\left[R-\left(\sqrt{R^2+z^2}-z\right)\right]}{z}<1$

Hence, option D is correct.