Connecting resistor in LED parallel circuit
The circuit you have shown is also not a very good idea. Ideally, each LED requires a series resistor. If the forward volt drop of one LED is a few percent smaller than the forward drop of another LED, the first LED will "hog" all the current and possibly burn-out. See this graph of forward volt drop and LED current: -



If one LED's characteristic curve starts at a slightly lower voltage (say 1.6 volts), it will be dropping 1.9 volts at 20mA and the other LED (20mA at 2volts) will be forced into taking a current of only 5mA because the terminal voltage of parallel LEDs is dictated by the LED with the smallest volt drop. What if a third LED was starting to conduct at 1.8 volts, with 1.9 volts across (as dictated by the first LED), this third LED will draw about 1mA.

In this simple scenario, you can see that the LED brightnesses will be varied a lot.

Ignoring this, for one resistor and one LED, if the current needed is 20mA and the forward volt drop of the LED is 2 volts then the resistor needs to "drop" a voltage that is \$V_{SUPPLY} - 2V\$. If the supply voltage is (say) 5V, the resistor carries 20mA whilst dropping 3 volts i.e. it has a resistance of 150 ohms.