Step 1: Start from the integrated rate law:
For a zero-order reaction the concentration falls linearly with time: $[\text{A}]_0 - [\text{A}] = kt$. The left side is a concentration, so $k = \frac{\text{concentration}}{\text{time}}$.
Step 2: Read off the unit:
Concentration is in $\text{mol dm}^{-3}$ and time is in $\text{s}$, so $k$ has the unit $\text{mol dm}^{-3}\text{s}^{-1}$. This matches the unit stated in the question.
Step 3: Test the other orders the same way:
A first-order law is $\ln\frac{[\text{A}]_0}{[\text{A}]} = kt$. The logarithm has no unit, so $k$ is in $\text{s}^{-1}$ and option (B) is wrong. For second order, $\frac{1}{[\text{A}]} - \frac{1}{[\text{A}]_0} = kt$, so $k$ is in $\text{dm}^3\text{mol}^{-1}\text{s}^{-1}$, and (C) is wrong. For third order, $k$ is in $\text{dm}^6\text{mol}^{-2}\text{s}^{-1}$, and (D) is wrong. Each step up in order multiplies the unit by $\text{dm}^3\text{mol}^{-1}$.
Final Answer:
Only the zero-order law gives $k$ in $\text{mol dm}^{-3}\text{s}^{-1}$, so the answer is option (A).
\[ \boxed{\text{Zero-order reaction (A)}} \]