I'd guess that the charged plane is supposed to extend indefinitely in all 'sideways' directions, so that the field it gives rise to is uniform and directed normally outwards from the plane. Gauss's law shows the magnitude of that field to be $$E_{\text{plane}}=\frac \sigma {2\epsilon_0}\ \ \ \ \ \ \ \ [\sigma = \text{charge per unit area of plane}]$$ The blue line on your graph shows the resultant field due to the point charge and to the charged plane. The condition for the resultant field everywhere to the right of the point charge to be directed to the right is simply $$\frac {q_{\text{point}}}{4\pi{\epsilon}_0 r_1^2}>\frac {\sigma}{2\epsilon_0}$$ in which $r_1$ is the distance from the point charge to the plane.

I'd guess that the charged plane is supposed to extend indefinitely in all 'sideways' directions, so that the field it gives rise to is uniform and directed normally outwards from the plane. Gauss's law shows the magnitude of that field to be $$E_{\text{plane}}=\frac \sigma {2\epsilon_0}\ \ \ \ \ \ \ \ [\sigma = \text{charge per unit area of plane}]$$ The blue line on your graph shows the resultant field due to the point charge and to the charged plane (except that the field should get indefinitely large next to the point charge). The condition for the resultant field everywhere to the right of the point charge to be directed to the right is simply $$\frac {q_{\text{point}}}{4\pi{\epsilon}_0 r_1^2}>\frac {\sigma}{2\epsilon_0}$$ in which $r_1$ is the distance from the point charge to the plane.