Answer
$a_g = 9.82~m/s^2$
Work Step by Step
In part (a), we found that at $r_0$: $~~~~a_g = \frac{0.2495~c^4}{G~M_h}$
We can find $a_g$ for this very massive black hole:
$a_g = \frac{0.2495~c^4}{G~M_h}$
$a_g = \frac{(0.2495)~(3.0\times 10^8~m/s)^4}{(6.67\times 10^{-11}~N~m^2/kg^2)~(1.55\times 10^{12})(1.99\times 10^{30}~kg)}$
$a_g = 9.82~m/s^2$
