Answer
5.8$\times10^{6}$m
Work Step by Step
Time period, $T = (2.4$\times$3600) s = 8640 s$
r = $8.0\times10^{6}$ m
Using orbital period equation, $T^{2}=\frac{4π^{2}}{GM}\times
r^{3}$, we get M= $\frac{4π^{2}r^{3}}{T^{2}G}$
i.e , M= $\frac{4π^{2}\times(8.0\times10^{6}m)^{3}}{(8640s)^{2}\times6.67\times10^{-11}N\frac{m^{2}}{kg^{2}}}$ $\approx 4.1\times10^{24}$ kg
Gravitational acceleration, $a_{g}$ =$8.0 m/s^{2}$ (given)
We know, $a_{g} = \frac{GM}{R^{2}}$
Or
R= $\sqrt \frac{GM}{a_{g}}$ = $\sqrt \frac{6.67\times10^{-11}N\frac{m^{2}}{kg^{2}}\times(4.1\times 10^{24} kg)}{8.0\frac{m}{s^{2}}}$ = $5.8\times10^{6}$ m
