LED driver circuit help needed
Yes. Your application of the formula is correct. I can provide part numbers for better LEDs than that. "Soon" ... (just stumbled out of bed to complete a report)

At about 15 mW a 'good' modern LED will give >= 150 l/W or about 2.25 lumen. That's enough to illuminate 1+ sheet of A3 paper to good readable level with colour well discerned. If illuminating a small enclosed space it should be more than you need. 

The formula (from above) is

$$ P = {V_{in}^2 \over L \cdot 10^-6} $$

or

$$ L = {V_{in}^2 \over P \cdot 10^{-6}} $$

\$L\$ in henries, or remove \$10^{-6}\$ and use \$L\$ in μH.

With this sort of IC the brightness varies approximately with the square of the input voltage so you get a power ratio of 9:4 for voltage changing from 3V to 2V.


A 2:1 difference is hardly noticeable to the eye if you look at one light at a time.
Look at two side by side and you can distinguish down to the 20% - 50% difference range.
"Wall wash" with two LEDs side by side and you can typically see differences down to 10%.


LED: Look at Cree ML-E to start. There are much better now. More on that soonish.