Calculating the gauge of wire required for the secondary in a Tesla Coil
It should be fairly obvious that you need to lay 1200 wires side-by-side in a space of 13.5 cm, so each wire (including its insulation) needs to be no wider than

$$\frac{13.5 \text{ cm}}{1200} = 0.1125 \text{ mm}$$

This would correspond to roughly AWG 38 with about 6 microns of enamel coating. I have no idea what the actual thickness of magnet wire coatings are, so you may need to adjust your figures to suit. Some online sources suggest that it's on the order of 0.0002 in., which would be about 5 microns.