1. Assume the rate law

For the reaction $2A + B \to \text{products}$ let the rate law be $r = k[A]^m[B]^n$ where $m$ and $n$ are the orders with respect to $A$ and $B$.

2. Use the information about $B$

It is given that when concentration of $B$ alone was doubled, the half-life did not change.

For a reactant to have no effect on half-life here, the reaction must be zero order with respect to that reactant in the rate law. Thus, $n = 0$

So the rate law becomes $r = k[A]^m$

3. Use the information about $A$

When concentration of $A$ alone was doubled, the rate increased by two times. That means $2^m = 2$ So, $m = 1$

Hence the rate law is $r = k[A]$ This is an overall first-order reaction.

4. Find the unit of $k$

Rate has units: $\text{rate} = \text{mol L}^{-1}\text{s}^{-1}$

From $r = k[A]$ we get $[k] = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}$

5. Match with options

$\boxed{\text{s}^{-1}}$ So the correct option is D.

6. Compare with stored answer

Stored correct answer is A: $\text{L mol}^{-1}\text{s}^{-1}$.

But from the given kinetic data, the reaction is first order overall, so the unit must be $\boxed{\text{s}^{-1}}$ Therefore, I do not agree with the stored answer.