How to calculate Snubber for different load value
One of the problema with RC snubbers is that they leak current. This doesn't really matter if the leakage current is much lower than the load requires to work, but otherwise it does. In the circuit diagram in the question there is a 10nF capacitor in series with the load. If we neglect the load and the series resistor, at 50Hz this capacitor acts as a 

\$X_{capacitor}=\dfrac{1}{2\pi f C} = \dfrac{1}{2\pi × 50 × 10\cdot 10^{-9}} \approx 300\text{k}Ω\$ 

impedance which in series with your 5W bulb: 

\$R_{load} = \dfrac{U^2}{P} = \dfrac{230^2}{5} \approx 10 \text{k}Ω\$ 

apparently allows enough current to make it glow. It actually is a bit more complicated than that, with phase angles and all, but we're talking order magnitude to get a feeling for what is happening. Therefore it is important to size the snubber for the attached load.


  
    Notice: This also implies that a snubber in parallel with a relay switch does not fully switch off the load and leathal voltages may be present even when switched off!
  


I did find rule of thumb design guideline in a book, but I have no practical experience with these guidelines:

Design guidelines:

Resistor:


1Ω per Volt power supply rail voltage;
Sufficient power rating (eg. 2W).


Capacitor:


0.1μF per Ampère power supply current;
Voltage rating in accordance with power supply voltage.


An unrelated practical example:

I once built a snubber network 22nF/630V and 100Ω/0.5W for switching a small PC amplifier rated at 240V(AC) and 70mA. That configuration did solve the problem I was facing back then even though it doesn't exactly follow the above design guidelines.