Answer
The positron will emerge from the field at $~~x = 0$
Work Step by Step
We can find the change in electric potential energy if the positron moves from $V=0$ to $V = 500~V$:
$\Delta U = q~\Delta V$
$\Delta U = (1.6\times 10^{-19}~C)~(500~V-0)$
$\Delta U = 8.00\times 10^{-17}~J$
In order to reach a location with an electric potential of $500~V$ the initial kinetic energy of the positron must be at least $8.00\times 10^{-17}~J$
We can find the initial kinetic energy:
$K_i = \frac{1}{2}mv_i^2$
$K_i = \frac{1}{2}(9.109\times 10^{-31}~kg)(1.0\times 10^7~m/s)^2$
$K_i = 4.55\times 10^{-17}~J$
The positron does not have enough mechanical energy to reach a location with an electric potential of $500~V$
Therefore, the positron will emerge from the field at $~~x = 0$