Answer
$a_g = 9.83~m/s^2$
Work Step by Step
To find the gravitational acceleration at the surface, we can assume that all of the mass is concentrated at the center of the Earth:
$a_g = \frac{GM}{R^2}$
$a_g = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)}{(6.37\times 10^6~m)^2}$
$a_g = 9.83~m/s^2$
