Your circuit has 24 parallel circuits containing 2 LEDs and a 150 ohm current-limiting resistor.
A typical voltage drop of a red LED is approximately 1.7V, thus the two LEDs drop 3.4V and the resistor must drop the remaining 5.6V.
5.6V across 150 ohms gives us 37.33mA of current and consumes 209mW.
First of all, 37.33mA is running the LEDs a little hot. Running more current through a LED generally increases the brightness -- to point. The difference in brightness between 20mA and 30mA is marginal, methinks. So I would redesign the circuit to have 20mA or 25mA, unless you know that pumping up the current is going give you some advantage. Perhaps you could make a small test circuit and run various currents through the LEDs to see the minimum current for the brightness you desire.
Second, the purpose of the circuit is to generate light, yes? The LEDs in the circuit are producing the light and any power consumed by the resistor just produces heat. That being said, I would try to reduce the voltage (and thus the power consumption) of the resistor.
My recollection is that red LEDs consume about 180mW @ 20mA. My calcs are that each parallel leg in your drawing consumes 0.18 + 0.18 + .209 = 0.569mW. This gives the total circuit 0.569W x 24 = 14.6W.
So here is my suggestion:
Instead of 2 LEDs per leg, use 4. Thus more of the voltage that is applied to LEDs, the less is consumed by the resistor. In this configuration you drop 1.7 + 1.7 + 1.7 + 1.7 = 6.8V across the LEDs and only 2.2V across the resistor. If we reduce the current to the nominal 20mA then the resistor value would be 2.2V / 0.02A = 110 ohms. 2.2V across 110 ohms gives use a power consumption 44mW, down from your original 209mW.
While that might sound like too much, because you now have 12 parallel legs instead of the original 24 you get a nice power savings:
0.18 + 0.18 + 0.18 + 0.18 + 0.044 = 0.764mW per leg, making the whole circuit 9.168W which is 2/3 of the original circuit -- and reduces the part count by 12 resistors.
I have used generic 5mm red LEDs in the calcs. You would need to read the spec sheets of your LEDs and alter the resistor values for each color.
My caveat: it is 2:21AM here and I am running a fever, so double check anything I say!