Electronic – connect 500K resistor directly to the mains

The efficiency is:

$$\eta = \dfrac{p_{out}}{p_{in}} = \dfrac{V_{out}\cdot I_{out}}{V_{in}\cdot I_{in}}$$

For a linear regulator, assuming the output current is much greater than the regulator circuit's internal currents, we have \$I_{out} \approx I_{in}\$ thus:

$$\eta \approx \dfrac{V_{out}}{V_{in}}$$

Your calculation is correct, however you also have to consider the power rating of the resistor.

Power is \$I^2 \cdot R\$ which is 1040W. That's a physically huge and expensive resistor and your circuit would be wasting 95% of the energy that you put in before it even gets to the bulb.

Here's a typical resistor style capable of that level of power dissipation:

It's 300mm long, 60mm in diameter and costs more than $40 US.

Would it ever make sense? Possibly- if you needed a 1kW heater for some reason as well as the lamp then it could conceivably.

Another consideration is that the socket of the halogen lamp may not be designed to keep fingers away from the supply. No big deal with 12V- it's pretty hard to electrocute a human with 12V, but 220V mains could be lethal. Most approaches to supplying low voltage lamps provide galvanic isolation which protects the user.

A better choice would be a 220:12 transformer or an 'electronic transformer' which uses switching power supply techniques to reduce the voltage.

Electronic transformer:

For a hobbyist, an old PC power supply could provide regulated 12V at 5A without breaking a sweat and it should be completely safe. See articles on the internet on how to get it to turn on (you may have add a dummy load). It provides DC rather than AC, however halogen bulbs of that voltage and power don't care much- the life may be slightly reduced on DC.