Your schemaic isn't the solution, this is why:
The anode voltage can't be taken to a higher voltage than the microcontroller pin voltage - voltage dropped in the resistor - Vbe of the transistor. Asuming you are using a 5V supply, the pin will be say 4.5V, Vbe will be 0.6V and the voltage dropped in the resistor will be Ib *1K which you state is 0.1V so the anaode can't go higher than about (4.5 - 0.1 - 0.6) = 3.8V.
The cathode current has to flow through the logic gate output - the 74138 has been out of production for many years so I'm guessing you have a 74LS138 which has a maximum sink current of only 8mA. When running at 8mA, the output voltage will be in the region of 0.8V so the total voltage you can possibly get to drive the LED is 3V, and that's assuming the LS138 didn't die first.
Lets suppose the LS138 didn't die, the 3V is shared between the LED itself and the series resistor, 2.01V is lost in the LED so the remaining 0.99V is across the 22 Ohm resistor so the current can't be more than 45mA.
You need to get much higher current through the LED, especially as it is only on for a short time. In fact to some degree you can overdrive the LED because it has a cooling off period between your 800uS pulses although you haven't mentioned it's actual duty cycle.
The way to do it is the use a PNP transistor to drive the anodes. Connect it's emitter to 5V and it's collector to the LEDs, you will have to invert the drive signal from the microprocessor or the LEDs will be on when they should be off. Driving the anodes this way will not lose the 0.6V in the transistors B-E junction and when the micro pin is low, it will saturate the transistor, makng it an almost short circuit. Then do the opposite in the cathode side, connect an NPN transistor with it's emitter to ground and drive the base from the LS138 through a resistor. The collector goes throught the LED current limiting resistor to the LED anode. Now when the LED is switched on, it will have almost full 5V available to it and the current will only be limited by it's series resistor. Again the signal will have to be inverted so you may need to use something like a 74LS04 or another transistor to do this. If you use a 74LS04 you can use it's 6 inverters, one to each of your six signals.
So instead of a limit of 3V/45mA to the LED, you now have almost 5V (you have lost 2 x VCEsat in the transistors) and a series resistor of (5 - LEDVf) /0.08 ~36 Ohms will give you the 80mA you want. As a little overdriving is acceptable, I would suggest 33 Ohms giving a current of about 90mA.
Brian.