(B) The acceleration due to gravity,denoted by $g$,is not uniform across the Earth's surface.
Due to the Earth's rotation and its equatorial bulge,the radius of the Earth is greater at the equator than at the poles.
Since $g = GM/R^2$,where $R$ is the radius of the Earth,the value of $g$ is smaller at the equator and larger at the poles.
Using the equation of motion $h = (1/2)gt^2$,we get $t = \sqrt{2h/g}$.
Since $g$ is smaller at the equator,the time $t$ taken to reach the surface will be greater at the equator compared to the poles.
Therefore,the packets will not take the same time; the packet dropped at the equator will take a longer time to reach the surface.