Voltage/current conversion
Not without knowing something about the circuit. For example if it's a simple resistive load then it would be:
\$(12V \ \div 18V)^{2} \times 4.5W = 2W\$
\$2W \div 12V = 0.1666A\$
Or
\$ 4.5W \div 18V = 0.25A \$
\$ 18V \div 0.25A = 72\Omega\$
\$ 12V \div 72\Omega = 0.1666A\$
\$ 12V \times 0.1666A = 2W \$  

However, it's probably going to be more complex than the above. Depending on the circuit, it may be non-linear, draw the same or even more current at the lower voltage.
For another example take a switching regulator, in your example say you have a 100 percent efficient switcher (don't exist, but for easy calculations) delivering 5V (from the 18V input) to the rest of the circuit. 4.5W at 5V is 900mA. So the 18V at 250mA is turning into 5V at 900mA.
If we drop the input voltage to 12V, then to supply the required 4.5W the switcher will automatically draw more current at it's input in order to deliver the 4.5W at it's output. So 4.5W / 12V = 0.375A. The input voltage has dropped, but the current has actually  increased. This can be seen as a negative impedance at the switcher input.

With more details about the circuit a reasonable estimation could be arrived at, but I think the easiest way would be simply to run it from 12V and measure the current with a multimeter.