Electronic – arduino – Circuit to drive a Solid State Relay

This type of opto-triac is mostly used in mains voltage applications. Due to the limited current capabilities it's often used as a driver for a triac which is the actual switching device. Your requirements are modest, so you won't need that, and you can use the opto-triac to switch your load directly. The opto-triac is a cheaper solution than an electromechanical relay then, so at first sight looks like a better choice.
An important difference between electronic and electromechanical switches, however, is that the latter have a very low on-resistance, while the former always will have a voltage drop when switched on. That's the on-state voltage mentioned in the datasheet. This can be up to 3V, which in a 230V application won't matter much, but if your supply voltage is only 24V AC that's more than 10%. Your load will probably work at 21V, but you'll have to check it.

Repetitive peak off-state current is the leakage current when the triac is switched off. 2\$\mu\$A is a safe value.

Holding current is the minimum load current the triac needs to remain on when the gate is no longer driven. For an average triac your 20mA may be a bit low, but again the opto-triac's 3.5mA is a safe value. (Besides, the gate will be continuously driven, so it's a moot point. It is important in four-component dimmers, where the diac gives a pulse to switch on the triac, after which the triac is on its own.)

Then there is the minimum trigger current. That's the minimum current you have to supply to the LED to switch the triac on, and we'll have to calculate the series resistor accordingly.
Where did you get that 38\$\Omega\$ resistor value? You need figures 3 and 4 to calculate the value for the LED resistor. Figure 4 shows that 10mA is a safe value, and figure 3 shows that at 10mA the LED voltage will be maximum 1.3V. So \$R=\frac{3.3V - 1.3V}{10mA}=200\Omega\$ maximum. Your 38\$\Omega\$ would result in more than 50mA, which is not only more than Absolute Maximum Ratings (page 4), but also more than your microcontroller will be able to supply. So don't exaggerate, and pick a 180 \$\Omega\$ resistor. At lower resistances the current may become too much for your microcontroller's output. If you want more current through the LED (no more than 20mA, never use the Absolute Maximum Ratings!) you may want to use a transistor. Since you'd need a lot of them, consider a driver IC like an ULN2803.

In conclusion I think this opto-triac is a good choice. Alternatively, you may have a look at the MOCxxx series, for instance the MOC3012 needs only half of the LED current, which your microcontroller would appreciate. It doesn't give a nominal value for triac current directly, but from maximum power dissipation (300mW) we can derive that this should be 100mA. (It says peak repetitive surge current is 1A, 120pps, 1ms pulse width.)

You can put a LED in series with the LED in the solid state relay. That will make the total voltage you need to drive the combination higher, but as long as you have the voltage available that's no problem.

Let's say you want to use a green LED to indicate the SSR is being driven on, and you have a 5V output to turn on the SSR. Green LEDs drop about 2.1V. You should be easily able to find one rated for 20mA, as most are. Probably the 20mA spec is a maximum for the SSR (for you to check), so let's say we want 10mA going thru the string. The LEDs will drop 1.3V + 2.1V = 3.4V, leaving 1.6V for the resistor. 1.6V / 10mA = 160Ω

So, put the 160Ω resistor, LED, and SSR input all in series. Now you can drive the whole thing with 5V to turn on the SSR and light the indicator, and 0V or open for off. When on, it will draw about 10mA.

Added:

In the description above the same current is flowing thru the indicator LED as the SSR LED. It looks like this SSR wants fairly large drive current of around 20mA. While you can certainly find a LED that can handle that, most "normal" and cheap LEDs are rated for 20mA max. They will also be plenty bright at much less current.

As Russell mentioned, you can put a resistor accross the LED to split the SSR drive current between the resistor and the LED. With a total of two resistors, you can set the SSR drive current and the external LED current separately. Russell made it sound a bit mysterious, but it is actually very easy. Here is a circuit:

Let's say we're using a common green LED for the external indicator. As I said above, figure about 2.1V drop for one of those. The max drop of the SSR LED is 1.4V, so that leaves 5V - 2.1V - 1.4V = 1.5V accross R2. This resistor sets the SSR current, which we want to make 20mA in this example. 1.5V / 20mA = 75Ω

D1 will have about 2.1V accross it roughly independent of the actual current we put thru it. R1 will therefore pass 2.1V / 150Ω = 14mA. That's how much of the total current will NOT run thru D1. The D1 current will therefore be about 6mA, which is plenty to light up most normal LEDs for indicator use.

Anyway, the point is not these particular values, but how you can use two resistors to set the currents thru both LEDs independently but still predictably enough for this application.

Note that you can use the same idea if you want the indicator LED to take the higher current. In that case you put the parallel resistor over the SSR's LED. This could be useful if your indicator LED has to be visible in daylight and the SSR's LED has limited current capability.