As former electronics engineer, let me mention a couple of observations (before reading ihart's and papajam's comments, both of which I agree with):
1. The circuit board list "Interbalco B930/B940/B945." A user manual (in PDF format) for this type of equipment (B930/B940) is available online here
. The user manual mentions on p.3 that "The B930/B940 can be powered by any kind or 12volt battery" and gives instructions for the wiring by mentioning specifically that "either polarity can be used". In other words, any 12V DC power supply capable of supplying 1.5 Amps current will do.
2. The "block" lists 8.5V and 1.5 Amps, which translates to 12.75 Watts power. Since the "block" does not list the output voltage as DC, it must be AC voltage -- and therefore, the "block" is probably a simple transformer, not a power supply. Now, when AC gets converted to DC, the AC voltage needs to be converted from "effective voltage" to "peak-to-peak voltage" by multiplying the effective voltage by the square root of 2 (2^0.5 = 1.4142). Applied to 8.5 V AC, the resulting DC voltage would be 8.5 V AC * 1.4142 = 12.02 Volts DC. This calculation would confirm that (a) the "block" is a transformer, and (b) that a 12 V DC power supply (instead of a 8.5 V AC supply) should work as well.
It appears the input voltage level (8.5 or 9 Volts AC, or 12 Volts DC) is not critical. I'm pretty sure the Jameco 9 V AC power supply would work (but if you have an unused 12 V DC power supply lying around, I'd try that first).