Answer
$F = m~(2.96\times 10^{-7})~N$
Work Step by Step
We can find $a_g$ at $r = 1.5~m$:
$a_g = \frac{GM}{r^2}$
$a_g = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(1.0\times 10^4~kg)}{(1.5~m)^2}$
$a_g = 2.96\times 10^{-7}~m/s^2$
We can find the magnitude of the gravitational force on the particle:
$F = m~a_g$
$F = m~(2.96\times 10^{-7})~N$
