Determining a suitable Power Supply for a Light
You have 3 times as many LEDs than resistors, so it makes sense that they are arranged in many  series strings of 3.
Assuming they are all running at the same current, then the 200&Omega; resistor strings will have the highest Vf. If we assume typical 20mA-30mA maximum LEDs, running at 10mA or so, and we have a blue or white colour which has around 3.2V drop at this current, then:

(3.2V * 3) + (200&Omega; * 10mA) = 11.6V

If it's 12mA then we have

(3.2V * 3) + (200&Omega; * 12mA) = 12V 

For the 470&amp;Omega and red LEDs:

(2.1V * 3) + (470&Omega; * 12mA) = ~11.94V 

So 12V looks like a reasonable bet.

According to this link, The Vf of the same colour can vary considerably between manufacturers for the same colour (it shows 6 white LEDs driven at 3.4V vary from 10mA to 44mA) so until you test yours you can only guess. 12V is a commonly available supply voltage though, so this looks plausible.
I would start ramping up the voltage slowly (ideally with a bench supply) whilst monitoring the voltage across one (or a few - one of each value) of the resistors using a multimeter (e.g. notch up a bit, test, etc).
Using Ohm's law as above, head for 10 - 15mA (so e.g. for the 470&Omega; and 10mA you are looking for a 470 * 10mA = 4.7V drop) and see what supply voltage you are at when you reach this level. Then you can make a good guess at the original voltage. 

Here's a table of Vf for various colours (from dangerous prototypes):





78L05

Another reason to assume not much higher than 12V is the drop for the 78L05. According to the datasheet, it has a thermal junction to ambient (&theta;ja) resistance of 150&deg;C/W. So if it were supplying 50mA (it can go up to 100mA) and the supply were something like 20V, then:

(20V - 5V) * 50mA = 0.75W

150&deg;C/W * 0.75W = 112.5 &deg;C rise above ambient. 

It's absolute maximum operating temperature is 150 so it would be very hot and have little ambient operating range. For 12V is more reasonable:

(12V - 5V) * 50mA = 0.35W

150&deg;C/W * 0.35W = 52.5 &deg;C rise above ambient. Much better.