#### Answer

The free-fall acceleration toward the sun at the distance of the earth's orbit is $5.9\times 10^{-3}~m/s^2$

#### Work Step by Step

Let $g_s$ be the free-fall acceleration toward the sun;
$g_s = \frac{G~M_s}{r^2}$
$g_s = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(2.0\times 10^{30}~kg)}{(1.50\times 10^{11}~m)^2}$
$g_s = 5.9\times 10^{-3}~m/s^2$
The free-fall acceleration toward the sun at the distance of the earth's orbit is $5.9\times 10^{-3}~m/s^2$.