I don't think that this is the place for such question which is rather a physics question but since is here I will try to answer using the few info given.
Looking at the size of a 5 W ceramic resistor I can just guess that either it is a big rod, either you have a small contact area with it
The iron has a thermal conductivity 10 times higher than ceramic
The iron has a specific heat capacity of 444J/kgC , that means that with 1W, even all energy goes to the iron rod, you need 444 seconds to heat on kg with one Celsius degree, for 5W you need 88s, for a 250g rod 20s. Only with one degree with perfect thermal contact and no cooling at all.
Also the contact area is to small and all the heat is absorbed by the rod and dissipated into the air to fast compared with the exposed area of the resistor.
To show my point see the following schematic:
simulate this circuit – Schematic created using CircuitLab
TRc1 and TRc2 are thermal resistance of the ceramic from the core to the rod surface / from the core to the exposed surface of the resistor where TRc1 > TRc1 because the contact area with the rod is much smaller
TRair 1 and TRair2 are the thermal resistance from the rod surface to the air / from the resistor surface to the air where TRair2 > TRair1 because the area of the rod is much bigger than the exposed area of the resistor.
TRrod is the thermal resistance of the rod, for iron ten times lower than ceramic.
So you will have a much better thermal conductivity from the ambient temperature to the rod contact point while also a better thermal conductivity from the heater core to the finger.
Now , what to do ?
Use a better suited heater , some that has a metal surface for attaching a heat sink where you can attach the rod like this one or use a MOSFET suited for a heat sink.
Put a thermal insulation on the the area not in contact with the rod to minimize the loses if you need energy calculations.
Make the rod with a large flat end to attach the heater for better thermal contact.
It would be useful to have a rough calculation of the power needed to heat the rod , my opinion is that 5W is not enough.