If the earth was smoothed out, even the ocean beds, so that the earth was like a smooth ball , and the water of the earth's sea bodies was placed on the earth , how deep would the water be on average.

I tried the calculation myself , please tell me if there is error you can see.

Numbers:

Volume of water: 1.3 billion km cubed

Average ocean depth: 3.79 km

Average earth radius: 6371 km

First , to calculate the Earth's radius by considering it's concave shapes , subtract average ocean depth from earth's mean radius.(The average earth radius might take into account mountains and if not then say if radius should be increased)

Smooth earth radius = 6731 - 3.79 = 6727.21 km

Now calculate earth's volume

smooth earth volume = 4/3 * pi * r cubed

= 4/3 * pi * 6727.21 km cubed

= 1.27524 x 10 to power 12 km cubed

Then add to that ocean water's volume so that you can figure out radius of sphere with water on it and subtract it from sphere's radius without water on it.

smooth earth with water volume = (1.27524 x 10 to power 12 km cubed) + (1.3 billion km cubed)

= 2.57524 x 10 to power 12 km cubed

Then find radius of smooth planet with water on it:

V = 4/3 * pi * r cubed

V /(4/3 * pi) = r cubed

(cube root)(V /(4/3 * pi)) = r

cube root (2.57524 / (4/3 * pi)) = r

r = 2.209 km

Is this right and are the numbers I used right?

Answers would be appreciated , thanks.