Oops I erased the previous post since I was wrong.
B.T.W. the resistor you choose is perfect. but change the 390 ohm to 300..330 ohm for 15 ma.
This is the correct calculation
The transistor gain Hfe gives you an expecting range. It is very good when you design analog device. In digital , you use the Hfe to figure out the minimum current you need to have. But in reality you always but more current because you want to drive the transistor full ON or OFF, not in the middle
Put even 10 times the current needed on the base is not an issue. Just need to calculate the power dissipation with
(IB * VBE) + (IC * VCE)
If you check the speck http://www.farnell.com/datasheets/11802.pdf
You use the transistor in ON/OFF way , you will need to look at the VCE saturation which tell you that it is typical at 0.09 V for 10 ma and 0.2V for 100ma.
When you calculate the resistor value for the led to get 15 ma. you need to consider the transistor drop also. which is usually 0.2V for low current. (It will vary and it could be up to 0.6V).
Normally I use the minimum Hfe value for my calculation. Just to verify.
Then to get 15ma it will be
For the resistor in serie with the LED
(6V - Led Voltage at 15ma - 0.2V transistor drop )/15 ma = ~ (6v- 1.175v -0.2)/15ma = 308 ohm => 300 ohm
For the resistor at the base
Minimum Base Current = 15ma / (100hfe) = 150 micro-amp.
Since you put a 10K at the base of the transistor , the resistor is serie will need to provide around 0.75V at the base for 15ma IC current. (speck figure 2).
I assume a 0.3V drop on GPIO.
Then the maximum resistor possible will be (3v -0.75v) / (( 0.75V /10K) + 150 ua) = 10000K
So the Maximum Resistor possible will be around 10K but something smaller will be safe.
Around 5K is a good choice.
B.T.W. Something wrong with your calculation according to the pictures.
You got 13ma and you have a led voltage to 1.175V then
(13 ma * 390 ohm) + 1.175V = 6.245 V ???? If you have 6V supply this is impossible. And you transistor will drop some voltage. When you add a current meter in the circuit, you change the system because current meter has a resistor value.
I suspect that you check the voltage on the led, add the meter in "milli-ampere" mode to read the current. This is wrong since the meter will add a resistor into the circuit. The best way is to check the voltage on the 390 ohm resistor and calculate the voltage.