In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by $$0.5$$ mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while $$46^{th}$$ division the circular scale coincide with the reference line. The diameter of the wire is ______ $$\times 10^{-2}$$ mm.

A screw gauge has 100 circular scale divisions and a pitch of 0.5 mm, meaning the main scale moves 0.5 mm per complete rotation. The least count is $$\frac{0.5}{100} = 0.005$$ mm.

Because the zero of the circular scale lies 6 divisions below the reference line, there is a positive zero error of $$+6 \times 0.005 = +0.03$$ mm.

When measuring the wire, the main scale reading is 4 divisions $$\times$$ 0.5 mm = 2.00 mm, and the circular scale reading is 46 $$\times$$ 0.005 = 0.23 mm. Thus the observed reading is 2.00 + 0.23 = 2.23 mm, and the corrected reading is 2.23 - 0.03 = 2.20 mm, which can also be written as $$= 220 \times 10^{-2}$$ mm.

Hence, the diameter of the wire is $$\boxed{220} \times 10^{-2}$$ mm.