Answer
It will not return to earth's vicinity.
Work Step by Step
We use the equation for escape velocity to obtain:
$v_{esc}=\sqrt{\frac{2GM}{R}}$
$v_{esc}=\sqrt{\frac{2(6.67\times10^{-11})(5.98\times10^{24})}{6,371,000}}\approx \fbox { 12 kilometers per second}$
This is greater than the escape velocity of the sun, so it will not return to earth's vicinity.
