How to determine watts and ohms for circuit board resistors?
The equation is basic Ohms Law with parts replaced for power calculation.

The formula is:
$$
P=\frac{V^2}{R}
$$
As it's got 12V across the resistor, and the resistor measures 51Ω, that becomes:
$$
P=\frac{12^2}{51} = 2.82W
$$
So you need at least 3W to be safe.

Placing resistors in parallel results in lower resistance.  The long form of the formula for that is:
$$
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n}
$$ 
So taking three 150Ω resistors in parallel gives you:
$$
\frac{1}{R_T} = \frac{1}{150} + \frac{1}{150} + \frac{1}{150}
$$
$$
\frac{1}{R_T} = 0.02
$$
$$
R_T = \frac{1}{0.02} = 50\Omega
$$
If all the resistors are the same value then it is a simple as dividing the value by the number of resistors, and the power ratings just add together, 20 three 0.5W resistors would give 1.5W of power.  Not enough for your application.

If you cannot get 1W resistors then you would be better with six 0.5W resistors of double the value - so 300Ω each.