Using NPN transistors as on/off switches for a LED circuit
Your basic concept is right, but there are some issues:

Paralleling 4 LEDs is not a great idea.  It would be better to give each LED its own resistor.  You say the LEDs drop 3 V.  Figure about 200 mV for the saturated transistor, which leaves 1.8 V across the resistor.  (1.8 V)/(20 mA) = 90 &Omega;.  The common value of 100 &Omega; would work fine.

You have to consider the gain of the transistor in determining its base resistor.  As you say, the transistor will pass 80 mA when on.  Let's say we want to run the transistor at 20:1 collector/base current, since it can be counted on to have more gain than that.  That means you want about 4 mA base current.  If the digital output goes to 5 V when high, you have about 4.3 V across the resistor.  That comes out to (4.3 V)/(4 mA) =  1.08 k&Omega;.  We have some slop in the calculations so 1.1 k&Omega; should work fine.

In any case, 10 k&Omega; is likely too high.  Working backwards, that provides 430 &micro;A base current, which means you are relying on the transistor to have a gain of at least 186.  That is well above the guaranteed minimum, so 10 k&Omega; is clearly too much resistance.