Step 1: Build a single conversion factor.
When the drainage coefficient $DC$ is in mm/day and the area $A$ is in $\text{km}^2$, the discharge in $\text{m}^3/\text{s}$ works out to $Q = \dfrac{DC \times A \times 1000}{86400} = \dfrac{DC \times A}{86.4}$.
This comes from $1\ \text{km}^2 = 10^6\ \text{m}^2$ and $1\ \text{mm} = 10^{-3}\ \text{m}$, so $DC \times A$ in mm.km$^2$/day scales to $\text{m}^3$/day by a factor of 1000, then to $\text{m}^3$/s by dividing by 86400 seconds.
Step 2: Substitute the given values directly.
$Q = \dfrac{25 \times 0.2}{86.4} = \dfrac{5}{86.4} = 0.0579\ \text{m}^3/\text{s}$.
Final Answer:
Rounded to three decimal places, the discharge is 0.058 cubic metres per second, the same figure the unit by unit method gives.
\[ \boxed{Q \approx 0.058\ \text{m}^3/\text{s}} \]