Solution:

Given parameters:
- Dimensions of the wooden pole: 25 mm x 25 mm x 4 m
- Length of the pole exposed: 0.6 m
- Specific gravity of sea water: 1.025
- Specific gravity of wood: 0.60
- Specific gravity of steel: 7.85

Buoyancy force on the wooden pole:
The buoyancy force on the wooden pole can be calculated using Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

The volume of the wooden pole submerged in water can be calculated as follows:

Volume of the wooden pole submerged = Cross-sectional area of the pole x Length of the pole submerged
= (25 x 25) mm^2 x 600 mm
= 375000 mm^3
= 0.375 m^3

The weight of the water displaced by the wooden pole can be calculated as follows:

Weight of water displaced = Volume of water displaced x Density of water
= 0.375 m^3 x 1025 kg/m^3
= 383.4375 kg

The buoyancy force on the wooden pole can be calculated as follows:

Buoyancy force = Weight of water displaced x Specific gravity of wood
= 383.4375 kg x 0.60
= 230.0625 kg

Weight of the steel attached:
The weight of the steel attached to the wooden pole can be calculated as follows:

Weight of the steel attached = Buoyancy force - Weight of the wooden pole
= Buoyancy force - (Volume of the wooden pole x Density of wood x Specific gravity of wood)
= 230.0625 kg - (25 mm x 25 mm x 4000 mm x 0.60 kg/m^3 x 0.60)
= 230.0625 kg - 13.5 kg
= 216.5625 kg

Therefore, the steel weight attached to the wooden pole is 216.5625 kg.

Conclusion:
The buoyancy force acting on the wooden pole is calculated using Archimedes' principle. The weight of the steel attached is calculated by subtracting the weight of the wooden pole from the buoyancy force. The specific gravity of sea water, wood and steel are used in the calculations as they determine the buoyancy and weight of the objects.