Full Bridge Rectifier with Filter Capacitor circuit
Vmax = Vavg(dc) + Vpp(ac)/2  is an approximation that the ripple is a triangle wave, but actually it is closer to a sawtooth.

You have estimated Vmax = 12.24V

Although the charge time is faster than the discharge time, the decay time and load current yields the peak-peak voltage AC ripple  ΔV=dV/dt*T  [Vpp] 


for interval T=1/2f and f=60Hz
for Ic=C*dV/dt in the capacitor discharge interval
from above ΔV = dV/dt*T 
thus ΔV=Ic*T/C  for T=1/120Hz [s], C=250uF and Ic = 30mA CC discharge only 


ΔV = 30mA*(1/120Hz) * (1/250uF) = 1 Vpp
let's get back to that ripple waveform

what ratio of each half cycle is used for charged/discharge time?
this ratio is directly related to  % ripple and charge/discharge current ratio


we know % ripple = ΔV /Vdc*100% is approx 1Vpp/12Vdc = 8.3%. 


If we double that we can reduce the ripple from 1Vpp to ~ 86% of 1V or 860 mVpp.  
This agrees well with simulation below. 405(+pk) + 445(-pk) = 850mV
With more proof I can show % V ripple is same as % load/charge current
I'll let you compare your proof to my estimate 


Thus charge current pulse in Diode and Cap= Ipeak = 30mA*1/ΔV[%] = 361 mA  
This agrees well with simulation below. (385 mA)


Note in the simulation below that the lower scope during power up is a couple Amps limited by the Diode ESR
Also consider that for ideal diodes and cap, if power is switched on at peak AC voltage is the diode and cap current is infinite !!



Food for thought