Step 1: Think of it as a small-change approximation:
If the radius grows by a tiny amount $dr$, the area grows by roughly a thin ring of circumference $2\pi r$ and width $dr$, so $dA \approx 2\pi r\, dr$, giving $dA/dr = 2\pi r$.
Step 2: Plug in the given radius:
At $r = 6$, $dA/dr = 2\pi(6) = 12\pi$.
Final Answer:
The required rate is $12\pi$ sq cm per cm.
\[ \boxed{12\pi} \]