a piece of wood of volume 200cm3 and density of 1.05gm/cm3. what volume of wood would remain above the surface of the liquid

None, sinks

That density is greater than fresh water, typo?

mass = 1.05 *200 = 210 gm

so Archimedes says mass of water displaced = 210 gm
but water(fresh, out of my geography) is 1.00 gm/cm^2
so
210 g = 1.00 * V of water
V of water = 210 cm^3 of water displaced, we need a hunk of styrofoam :)

As far as I know lignum vitae is still used for many stern tube bearings, although probably no longer by the Navy.

To determine the volume of wood that would remain above the surface of the liquid, we need to compare the density of the wood and the density of the liquid.

Given:
Volume of wood = 200 cm^3
Density of wood = 1.05 g/cm^3

Since we are not given the density of the liquid, we cannot directly calculate the remaining volume above the liquid surface. However, we can assume that the liquid is water, which has a density of approximately 1 g/cm^3.

To find the remaining volume above the liquid surface, we need to compare the densities of the wood and the liquid. If the density of the wood is greater, the wood will sink completely. If the density of the liquid is greater, there will be some portion of wood above the liquid surface.

Comparing the density of the wood (1.05 g/cm^3) with the assumed density of water (1 g/cm^3), we can see that the wood's density is slightly higher.

Therefore, in this scenario, the wood would likely sink completely below the surface of the liquid; there would be no remaining volume above the liquid surface.

I am reminded of the stern tube bearings on the battlship IOWA, when we pulled the shafts to reactivate. The stern tube bearings were made of water lubricated Lignum vitae( is a wood), also called guayacan or guaiacum. Look up the density of that wood. http://www.engineeringtoolbox.com/wood-density-d_40.html

We replaced all that wood with oil lubricated bearings. I still have a piece of that wood bearing on my desk.