Step 1: The volume under a triangular unit hydrograph represents exactly 1 cm of direct runoff spread over the whole catchment.
Step 2: Volume of a triangle $= \frac{1}{2} \times \text{base} \times \text{height}$, so here Volume $= \frac{1}{2} \times (144 \times 3600 \text{ s}) \times 23 \text{ m}^3/\text{s} = 5{,}961{,}600 \text{ m}^3$.
Step 3: Convert this volume to an equivalent depth of 1 cm ($0.01$ m) over an unknown area $A$: $A = \dfrac{5{,}961{,}600}{0.01} = 596{,}160{,}000 \text{ m}^2$.
Step 4: Converting to km$^2$ ($1 \text{ km}^2 = 10^6 \text{ m}^2$): $A = \dfrac{596{,}160{,}000}{10^6} = 596.16 \text{ km}^2$.
\[\boxed{A \approx 596 \text{ km}^2}\]