The answer would be right (approximately) if your LC concentration was 300 ppm, which it is not. However, I don't like how you've stated step 2, which is close to being meaningless. Dissolving a neat sample in solvent is not generally referred to as "dilution", and it would be impossible to quantify the "dilution" factor unless you know the volume of the neat sample, which I assume you don't. Work with volume, not "dilution".
The concentration of your LC sample is 196 ppm. Let's assume, as Corribus suggests, that this is ppm m/v, so 196 ppm = 0.196 mg/mL. So your 1 mL sample contains 0.196 mg analyte. This was all originally in the 5μL aliquot, so the concentration of this was 0.196 * 1000/5 = 39.2 mg/mL.
So the amount in your original solution was 39.2*V, where V is the volume of the solution. But we don't know this. If you add 1.487 g solute to 20 mL solvent, the solution volume will be significantly greater than 20 mL, but you don't know a priori exactly what it is. You could assume that the solute has a density of 1 g/mL, and that the volumes are additive. Or you could assume that what you actually did (and what you should have done) was to dissolve the sample in less than 20 mL ethanol, and make up the solution volume to 20 mL with ethanol. Or you could actually measure the solution volume, if you have accurate enough equipment, and this was a real experiment, not a homework question.
Let's assume, though, that your solution volume was 20 mL. The mass of analyte in this solution would be 39.2*20 = 784 mg. So the mass fraction (we wouldn't usually say "concentration") of analyte in your original sample was 784/1.487 = 527 mg/g.
This is what we would get following your method, but without the conceptually dodgy statement that "20 mL is 20 dilution". It also involves an assumption, which may be quite inaccurate, about the volume of the solution. I hope you appreciate why I've done it this way, and the nature of the assumptions involved.