Answer:
The weight would be greater in the Mariana because the radius of the earth is lower.
Step-by-step explanation:
We will make a comparison through constants and equations to see which one is more viable.
Using the force of gravity

F = Gravity force
G= Gravitational constant
m= mass
R is the radius of the earth,
km and
km
Density
is equal in both places,
-Gravity Force at Ecuador,

-Gravity Force at bottom of Mariana trench,

Making the relation,


For the Ecuador,

If we take
as a constant X then

So the gravity force of the place is inversely proportion of the radius
Same for the Mariana trench,

Then, the weight would be greater in the Mariana because the radius of the earth is lower.