Let me take a crack at this...
The 553-F44X is a 115V to 12.6 Vct (2 A) transformer.
For a 12.6 VAC RMS voltage source the peak voltage will be 12.6*1.414 = about 18V.
The 1n4007 will drop maybe 0.5V, so the peak will be say 17.5V.
Power (P) = V*I = V^2/R so now you can compute the (max) power dissipated through the resistor at any point in time.
(In this particular circuit it could be argued that you could get by with a much lower power rating because of the half-wave rectification performed by the diode and considering the average power dissipated by the resistor, but it is always safer to perform this calculation based on the extreme conditions.)
For the fuse...
Let's say the secondary side draws about 20 mA (peak). The transformer has about a 10-to-1 turns ratio, so that means 2 mA on the primary side, so it would appear that 1/16 A (about 62mA) would be plenty.
The problem is that you also have to account for inrush current when power is initially turned on. Probably a 1/16A fuse will suffice. And it will depend on whether you use a fast-acting or slo-blow fuse. You might just have to try it and see if you start blowing fuses.
Btw - if you are going to play around with bare transformers and mains voltages I would invest in a "Quicktest" connector:
https://www.cliffuk.co.uk/products/tools/quicktest.htmhttps://youtu.be/_DTmL73th7YIt is a much safer way to connect mains power to your circuits.