I'll bet that those lamps are just standard 2W 12V grain-of-wheat bulbs.

One can work it out based on the wiring diagram given which shows the 40W ballast resistors in series with 2x 2W lamps.

If you take the mains voltage of (say) 230V and multiply it by 2/44th (2+2+40W), you get a bit over 10V per bulb.

Going further, I bet that ballast resistor is likely to be either 1100 ohm or more likely 1200 ohms being a 'Prefered value'.

(40/44)*230V = 209V = voltage drop across ballast.

R = (V^2) / P

so (209*209)/40

43681 / 40

= 1092 ohms

The numbers work out slightly better if we assume 240V instead on 230V (closer to 1200 ohms).

It'll be interesting to see what a multimeter thinks that resistor is.

Anyway, as for where to get a bulb, try these Jaycar ones - SL2675 or SL2685

(Being new I can't post a link to it, so you'll have to use Jaycar's less than stellar web search function for those 2 part numbers)

They state they are 12V 0.1A so you'd need to wire 2 of them in parallel with each other so they don't get too much current through them and burn out.

The old lamps would be about 0.2A

I = P/V

230V / 44W = 0.191A = 191mA

Either that, or hunt around for some 2W, 12V grain of wheat lamps from somewhere.