I get about 11300 km radius water drop to get the same g as earth. So the preassurevat Challenger Deep depth would be the same?

Yes, your density to mass conversion is correct.

V,h2o [m^3 ] * rho,h2o [kg/m^3 ] = V,earth [m^3 ] * rho,earth [kg/m^3 ]

With the density of water, volume of earth, and density of earth constant, you can use this equality to find the required volume, which will be proportional to the ratio of densities.

As you have already done, multiplying volume by density yields mass, and with constant density, volume is directly proportional to mass.

Lots of words to say yes, you are correct in terms of volume required to give the same gravity given the above assumptions.

This next paragraph relies on many invalid assumptions: However, due to the different volume and identical gravity, the pressure would increase more slowly with depth than on earth. This is because water is in fact slightly compressible, and under such enormous forces it would in fact be more dense as you approached the core. If you took the density as an average density, which is inaccurate, you would see a pressure/depth change similar to how it is on earth, with a similar pressure to the earth's core at the core of your big water droplet, and a similar pressure at the depth of the challenger deep that you would see on earth.

Again, p = rho g h, so if g = G,earth, and rho = rho,h20, then p will increase linearly with the 'height' of the water above you