How can I calculate current(I) when I don't know voltage drop
Normally, you don't want a fixed voltage for your LED. You want a current. (\$20mA\$ not uncommonly.) So you are focused on the wrong goal.

But let's say you really do want to set up some fancy voltage regulator for your \$9V\$ battery. (Instead of doing the sensible thing about setting up the current, instead.) You can try this. Just use a \$100\Omega\$ resistor where the LED goes, at first, and adjust \$R_2\$ until you measure a voltage across the \$100\Omega\$ resistor that you like (as shown, where the (-) and (+) appear in the schematic and where you place your voltmeter for measuring.) Then you can replace it with the LED and see how that goes.







simulate this circuit &ndash; Schematic created using CircuitLab

You can play with the potentiometer, \$R_2\$, to then adjust the voltage at your LED. Should only waste an excess of \$2mA\$, regardless of the set voltage. The maximum output voltage will be \$6V\$ with a current compliance of up to \$100mA\$. (\$Q_4\$ may need to be a TO-220 packaged type, just in case.) The problem will be that you will have a very hard time adjusting the potentiometer without having a serious impact on the LED brightness. But it can be a learning experience.

Or you could just use a variable voltage regulator IC.

But you really should just shoot for setting a current using a simple resistor, using the formula \$R=\frac{9V - 1.9V}{20mA}= 355\Omega\$. Then go get a \$330\Omega\$ or a \$390\Omega\$ resistor. It will work just fine. Then measure the voltage across your LED, too. See where it is.