Answer
The particle reaches a distance of $~~2.5\times 10^5~m~~$ from the surface.
Work Step by Step
We can write a general expression for the gravitational acceleration at the surface:
$g = \frac{GM}{R^2}$
We can use conservation of energy to find the distance $d$ the particle reaches from the center of the asteroid:
$K_2+U_2 = K_1+U_1$
$0+U_2 = K_1+U_1$
$-\frac{GMm}{d} = \frac{1}{2}mv^2-\frac{GMm}{R}$
$\frac{1}{d} = \frac{GM}{GMR}-\frac{v^2}{2GM}$
$\frac{1}{d} = \frac{2GM-v^2~R}{2GMR}$
$d = \frac{2GMR}{2GM-v^2~R}$
$d = \frac{2gR^3}{2gR^2-v^2~R}$
$d = \frac{2gR^2}{2gR-v^2}$
$d = \frac{(2)(3.0~m/s^2)(5.0\times 10^5~m)^2}{(2)(3.0~m/s^2)(5.0\times 10^5~m)-(1000~m/s)^2}$
$d = 7.5\times 10^5~m$
We can find the distance $h$ above the surface:
$h = d-R$
$h = (7.5\times 10^5~m)-(5.0\times 10^5~m)$
$h = 2.5\times 10^5~m$
The particle reaches a distance of $~~2.5\times 10^5~m~~$ from the surface.
