The rate law for an elementary reaction where the concentration of reactant M impacts the rate of the reaction is expressed as:

$$\text{Rate} = k[M]^n$$

where k is the rate constant, M is the concentration of the reactant, and n is the order of the reaction with respect to M.

In the scenario described, when the concentration of M is doubled, the rate of disappearance of M increases by a factor of 8. Let’s denote the initial concentration of M by [M] and the doubled concentration by 2[M]. According to the rate law, the initial rate would be:

$$\text{Rate}_1 = k[M]^n$$

Upon doubling M:

$$\text{Rate}_2 = k(2[M])^n = k2^n[M]^n$$

It is given that this rate is 8 times the initial rate:

$$ \text{Rate}_2 = 8 \times \text{Rate}_1 $$

$$ k2^n[M]^n = 8k[M]^n $$

Cancelling out the common terms and simplifying the equation, we have:

$$ 2^n = 8 $$

To find n, solve for n in:

$$ 2^n = 2^3 $$

Hence, n = 3, indicating that the order of the reaction with respect to M is 3. The correct option is therefore:

Option B: 3