Answer
$a_g = 1.3\times 10^{12}~m/s^2$
Work Step by Step
We can find the gravitational acceleration at the surface:
$a_g = \frac{GM}{R^2}$
$a_g = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(2.0\times 10^{30}~kg)}{(10^4~m)^2}$
$a_g = 1.3\times 10^{12}~m/s^2$
