Part a) From the force balance, you know that ρsVsg = ρw(1/3)Vsg , so the solid (s) density is 1/3 that of water.
From the definition of density, you know that ρs = ms/Vs or that Vs = 3kg / ρs in m3.
From the equation of volume of a cylinder V= πr2h = (1/4)πd2h you can substitute d/5 for h and solve for d in meters which I leave to you.
Part b) I'm assuming that the course is calculus-based because the bouyancy force is not constant with depth submerged:
The change of energy going from the equilibrium position to the submerged position (The Min W required)
ΔE = Int(from 1/3 h to h) of (A(P(z)-Patm)dz) - ρVgΔz where z is the depth from the surface.
ΔE = Aρwg * Int(from h/3 to h) of (zdz) - msgz = (πd2/4)ρw g(z2/2) from h/3 to h - msg(2/3 h)
***ΔE = (πd2/4)ρwg(8h2/18) - msg(2/3 h) (BTW, use 1000 kg/m3 for density of water.)
A way to do this without the integral is to realize that the value of a linearly changing force has an average equal to the average of the endpoints. This way the energy change due to work against the buoyant force will be FavgΔz or 1/2 (F(h) + F(h/3)) (2h/3) = (1/2)Aρwg(h+h/3)(2/3 h) = Aρwg(8h2/18) = Aρwg(4/9)h2
I guess easier still is multiplying F(2/3 h)(h-h/3). These all give the same answer for the integral term.
b) I will leave this to you. The work will be due to an increase in elevation - the average bouyant force times the difference in height.