Solution:

Combining two resistors in parallel:

When two resistors are connected in parallel, their combined resistance is given by:

1/R = 1/R1 + 1/R2
R = R1R2/(R1 + R2)

Given that the resistors have a resistance value of 2000Ω each, we can substitute R1 = R2 = 2000Ω in the above equation to get:

R = 2000 x 2000 / (2000 + 2000)
R = 1000Ω

Therefore, the combined resistance of the two resistors in parallel is 1000Ω.

Calculating the wattage rating:

When resistors are connected in parallel, the total power dissipated is the sum of the power dissipated by each resistor.

Power dissipated by a resistor is given by:

P = V^2/R
where P is power in watts, V is voltage in volts, and R is resistance in ohms.

Since the two resistors are connected in parallel, they have the same voltage across them. Let's assume this voltage to be V.

The power dissipated by the first resistor is given by:

P1 = V^2/R1
Substituting R1 = 2000Ω, we get:
P1 = V^2/2000

Similarly, the power dissipated by the second resistor is given by:

P2 = V^2/R2
Substituting R2 = 2000Ω, we get:
P2 = V^2/2000

The total power dissipated by the two resistors in parallel is given by:

PTotal = P1 + P2
PTotal = V^2/2000 + V^2/2000
PTotal = 2V^2/2000
PTotal = V^2/1000

Since we know that the two resistors have a total power rating of 2 watts, we can equate this to the total power dissipated by the two resistors:

2 = V^2/1000
V^2 = 2000
V = sqrt(2000)
V = 44.72 volts (approx.)

Substituting this value of V in the expression for PTotal, we get:

PTotal = V^2/1000
PTotal = (44.72)^2/1000
PTotal = 2 watts

Therefore, the combined resistance value and wattage rating of the two resistors in parallel are:

Resistance value = 1000Ω
Wattage rating = 2 watts

Hence, option B is the correct answer.