Please help?

Take g = 9.81 m/s² = 9.81 N/kg
Gravitational force acting on the piece of metal = mg = (3.6 kg) × 9.81 (N kg⁻¹) = 35.3 N
Volume of the metal = (3.6 kg) / (9000 kg m⁻³) = 0.0004 m³

(a)
The density of air is very small, and thus the upthrust acting on the metal is negligible.
Reading of balance = 35.3 N ≈ 35 N (to 2 sig. fig.)

(b)
Density of water = 1000 kg m⁻³
Upthrust acting on the piece of metal = (0.0004 m³) × (1000 kg m⁻³) × (9.81 N/kg) = N
Reading of balance = (35.3 - 3.9) N = 31.4 N ≈ 31 N (to 2 sig. fig.)

(c)
Upthrust acting on the piece of metal = (0.0004 m³) × (1200 kg m⁻³) × (9.81 N/kg) = 4.7 N
Reading of balance = (35.3 - 4.7) N = 30.6 N ≈ 31 N (to 2 sig. fig.)

(d)
Mass of the metal to 2 sig. fig. = 35 N
Upthrust acting on the piece of the metal = (35 - 33) N = 2.3 N ≈ 2 N
Density of the liquid = (2 N) / [(9.81 N kg⁻¹) × (0.0004 m³)] = 510 kg m⁻³

volume V = m/ρm = 3.6 kg /9.0 kg/dm^3 = 0,400 dm^3 ; g rounded up to 10 m/sec^2

..what is the reading of the balance in N :
(a) with the metal in air
air density ρa = 0.00125 kg/dm^3
reading R1 = m*g*(ρm-ρa) = 3.6*10*(9-0.00125) = 324 N

(b) with the metal in water
water density ρw = 1.00 kg/dm^3
reading R2 = m*g*(ρm-ρw) = 3.6*10*(9-1) = 288 N

(c)with the metal in brine
brine density ρb = 1.20 kg/dm^3
reading R3 = m*g*(ρm-ρb) = 3.6*10*(9-1.2) = 280.8 N

(d) what is the density ρℓ of a liquid if a reading of 33N is achieved?
33 = 3.6*g(9-ρℓ)
36*ρℓ = 36*9-33
ρℓ = 291/36 = 8.083 kg/dm^3