The venerable old 1N4148 may be a choice here.

I had taken the 1N4002 from another circuit using TDA2003, which is able to deliver a current of up to 1A. In an other design also containing TDA2003 even 2A can be delivered, and this with 1N5401! But TDA2003 is much slowlier than NE555, means its edges are smoother, so that the reverse recovery time of diodes will not cause a too high discharge for the small period when diodes had not already turned-off.

As NE555 shows much faster edges, a 1N4148 might indeed be a better choice!

The effect of finite reverse recovery time of diode is, that in the moment of toggling of output the formely charged capacitor is discharged a bit again, because diode is still turned-on.

The amount of this discharge can easily be estimated:
Assume that reverse recovery time of 1N4002 is about 1µsec. From datasheet of NE555 it can be seen, that voltage drop across internal output transistor is 1.7V when delivering 100mA. This yields an output impedance of 17Ohm. This is the resistance by which the 10µF capacitor is discharged again for about 1µsec. The result is a discharge of capacitor by about 30mV. This voltage fluctuation is the 'no load current ripple', which is relevant but acceptable for most cases, if adequate RC filtering is used, as discussed in my last post.

The use of 1N4148 would almost eliminate this 'no load current ripple', because reverse recovery time is in the nanosecond range.

If the exact formula of frequency of NE555 is not known, frequency can at least be estimated: A period is formed by a charge and discharge sequence of timing parts 18kOhm and 10nF, which yields 1 / (2 x 18kOhm x 10nF) = 2.8kHz. This is about 3kHz.
Correct formula gives: f = 1.44 / ((3.3kOhm + 2 x 18kOhm) x 10nF) = 3.7kHz

As period time for a 3.7kHz square wave is 270µsec, reverse recovery time of 1µsec of diodes is not relevant, means 1N4002 can be used at such low frequencies, but of course ripple is a bit increased.

Kai