[MrFixr55]

One way of dealing with this is to "reverse engineer" the Radiola 18 and PS.  With the "good" unit al hooked up, CAREFULLY measure your B+ voltages, both at the plates and cathode of the 80 Rectifier, then at the output of the filter, thereby determining the voltage drop across the inductors.  Even though these sets do not draw a lot of current, the "unloaded" output voltage of a power supply can be quite a bit higher than the output when the radio is connected to the power supply.  Another consideration is that although the no-load rectifier plate winding value of the original Radiola transformer and your new transformer are the same, the AC voltage for the rectifier winding on the Radiola transformer may drop when the various filament windings are under their normal load.

Another way of dealing with this is to determine the total current and specified B+ output. Test your power supply using a resistor across the B + to ground that will duplicate that load and measure your power supply's loaded output voltage.  Compare that to what you get for the values that you get for the complete original Radiola 18.  If my calculations were correct in my previous, the total B+ current load is 42mA.  So if E= IR and R= E/I, then R= 165 / 0.042= 3,929 Ohms.  The current capacity of the resistor needs to be sufficient.  So if P=EI, then 165 X 0.042 = 6.9W.  You may find that your supply puts out much more than 165V, even under this load. you may need to vary the resistance to find out the true output under a load of 0.042A

Next, subtract 165V from your measured "under load" output.  For example, if the under-load output of your power supply is 250V and the desired voltage is 165V and the current load is 0.042A (42mA), then the dropping resister needed is based on the following calculations:
250V (New PS output voltage) - 165V (Voltage required by Radiola 18 chassis) = 85V (dropping resistor voltage).  If E= IR then R= E/I and if the voltage drop is 85V and the current (which is constant anywhere in a series circuit) is 42mA or 0.042A, so 85/0.042 = 2,023 Ohm, or 2KOhm.  to further complicate things, it is good to put a filter cap after the dropping resistor.  This may increase current.  I will leave it up to the engineers to determine the additional current added by a cap.  I would assume that for a cap as low as 1uF, it would be low.  For pure DC, assuming leakage of the cap being none, then after charge, the current is 0A.  However, this is the real world, not the theoretical world.

When filtering pulsating DC, the lower the size of the filter capacitors, the lower the output but the higher the hum.  The inductors must be of a higher inductance (and possibly resistance) to compensate for this.  Early AC radios had capacitors of lower value than newer sets because electrolytic caps were not in use at the time.  Note that there are 3 inductors in the power supply.  2 are for B+ and the third is for the plate of the output tube.

In addition to the plate voltages for the various stages, check the "bias voltages", especially for the output, and verify the correct speaker wiring (see further below).

Here is the schematic of the output and power supply stages of the Radiola 18 (The 60 would be similar but not exactly the same):
   

In my first experiments with a Radiola 60 chassis and power supply found when I was 14 at the town landfill (since lost with the sale of the family home), I did not have a schematic, so I merely traced the wiring and connected an AA5 speaker between the 71A's plate and the output of the power supply.  The radio played loud but with objectionable hum.  I did not know that the 2 pin jacks on the power supply were for the speaker, since why would the speaker jacks be on the power supply chassis?  I did not have access to the Radiola 60 schematic? (There was no internet then and it was not in the local library.)  And, of course, the capacitive coupling would have a rather high cutoff at 60 Hz (the frequency of the filament-based ripple), and the bass response of the 100A and 103 speakers was not large to begin with.  These factors would mitigate hum.

Since the output stage is a "Class A", and since the speaker is capacitively coupled, I don't see a lot of harm done in operating the set without the speaker. However, if you are using a "more modern" speaker scavenged from a late 1940s to early 1960s 5 tube AM radio (that uses a 35L6, 50L6, 50C5, etc.), ensure that you are using the output transformer from that set.  Ensure that the "blue" lead is connected to the output coupling capacitor (which should be at least 0.5uF).  Ensure that the red lead is connected to the center tap of the resistor across the 5V filament supply to the 71A output tube.  If it is connected to B+ or if a 2K Ohm output transformer is not being used, then volume will be very low.