pressure = Integral[h=x to infinity] of rho g(h) dh where x is the current distance from the center. "h" here means "height." Since we're at the center, x=0. Since rho(vacuum) = 0, this reduces to Integral[h=0:200km] rho g(h) dh. I simplify and say that water is incompressible, with density of rho = 1000 kg/m
mm. (At 1 giga pascals the density is 1.2 that of its normal density.)
g(r) = G * M(r) / r * r where M(r) is the mass inside a sphere of radius r. That's G (rho 4/3 pi r * r * r) / r * r or 4/3 rho G pi r.
Therefore, pressure = Integral[h=0:200km] rho * (4/3 rho G pi h) dh. This is 2/3 rho * rho G pi h * h, giving 5.6 mega pascals.
Looking at the phase diagram you pointed out, 6MPa is well within the liquid range. (Presumably the temperature would be around 4C). Even with a higher density it's at most about 7MPa, and you need to get to 1 GPa for water to always be a solid.
In other words, unless I did my math wrong or left something out of the calculations, this water sphere isn't close to one of the ice phases of water.
BTW, for more fun, since there's no heat source for the water (radioactive decay, heat from the phase change to ice, etc), then how does it stay liquid? If exposed to vacuum it would turn to vapor until it cooled down enough (about -60C) to freeze. There isn't enough mass to hold down much atmosphere, so I presume it's covered.
How also does it get enough energy to stay liquid? Most of the energy will be dumped into the top layer, so there will be a warm zone on top of a thermocline, like with Earth's oceans. If the energy comes from the sun, then do the poles freeze? If it doesn't rotate fast enough then the backside will freeze.
If ice does form, it reflects more heat than water so that region will stay ice.