Automatic Solar light far too dim, help with increasing brightness
R2 sets the current through the LEDs. It must soak up the voltage difference between the supply and the LEDs. A typical 5mm super-bright white LED drops about 3.3V at 20mA. A 3.6V AA NiMH battery should deliver between 4.2V and 3.3V depending on state of charge, with an average of about 3.8V. The transistor will drop about 0.1V in saturation, leaving ~0.4V across R2. With 3 LEDs consuming 20mA each the total current is 60mA, so the resistance required is 0.4V/0.06A = 6.7 Ohms (a 6.8&ohm; resistor will be close enough). If you want a lower LED current then insert that into the formula.

If the LEDs are not well matched you may find that they do not draw equal current in parallel. This issue can be avoided by splitting R2 into 3 resistors which feed each LED separately. The value of each resistor must then be 3 times higher than the total, eg. 20&ohm; (use the nearest 5% value, 18&ohm; or 22&ohm;).

R1 sets the Base current of Q1. It should supply sufficient current to saturate the transistor, so that it can switch full battery voltage to the output at the required current. Collector current is amplified by the 2N3906's current gain (Beta or Hfe) which in normal operation is between 60 and 300. However in saturation the current gain is reduced, so a lower Collector/Base current ratio may be required (eg. 20:1). 

The value of R1 can be calculated in the same manner as for R2 (determine voltage across resistor, apply Ohm's Law). However that reveals a problem with your circuit. A 6V solar panel usually has 12 cells in series, supplying ~0.5V each. In the dark each cell acts like a silicon diode, so the panel is effectively 12 diodes in series. When powered externally the diode action will drop ~0.5V per cell, for a total of 6V. This is higher than the NiMH battery voltage, but your circuit needs a lower voltage (preferably zero Volts) to turn on the transistor. It only works due to leakage current in the solar panel, which is not defined and drops to zero as the panel voltage approaches zero. 

If you try to increase Q1's Base current by reducing the value of R1 then you may be limited by the panel's dark current. To guarantee sufficient Base current you can wire a resistor across the panel. The combined value of  this resistor (let's call it R3) and R1 then sets Q1's Base current. Assuming 3.8V from the battery, 0.6V across Q1 Emitter-Base, and 60mA/20 = 3mA Base current, R1+R3 = (3.8V-0.6V)/3mA = 1067&ohm;. Splitting that value into equal resistances gives 534&ohm; per resistor (470&ohm; will be close enough). R3 will waste about 12mA of potential charging current, but this should not be significant.

In bright sunlight the panel can deliver up to 330mA charging current. This may be slightly above the recommended trickle current for AA cells (eg. 200mA for a 2000mAh cell) however this high current is unlikely to cause damage as it will only occur for an hour or so in the middle of the day.