#### Answer

(a) The free-fall acceleration at the surface of the sun is $274~m/s^2$
(b) The sun's free-fall acceleration at the distance of the earth is $5.90\times 10^{-3}~m/s^2$

#### Work Step by Step

(a) We can find the free-fall acceleration $g_s$ at the surface of the sun.
$g_s = \frac{G~M_s}{R^2}$
$g_s = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg)}{(6.96\times 10^8~m)^2}$
$g_s = 274~m/s^2$
The free-fall acceleration at the surface of the sun is $274~m/s^2$.
(b) We can find the sun's free-fall acceleration $g_s'$ at the distance of the earth.
$g_s' = \frac{G~M_s}{R^2}$
$g_s' = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg)}{(1.50\times 10^{11}~m)^2}$
$g_s' = 5.90\times 10^{-3}~m/s^2$
The sun's free-fall acceleration at the distance of the earth is $5.90\times 10^{-3}~m/s^2$.