A thin, uniform rod has mass $m$ and length $L$. Let the acceleration due to gravity be $g$. Let the rotational inertia of the rod about its center be $md^2$.

Find the ratio $L/d$.

$ \textbf{(A)}\ 3\sqrt{2}\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ \text{none of the above} $
1. The problem states that the rotational inertia of the rod about its center is given by \( md^2 \). We need to find the ratio \( \frac{L}{d} \).

2. It is well-known that the rotational inertia \( I \) of a thin, uniform rod of mass \( m \) and length \( L \) about its center is given by:
   \[
   I = \frac{1}{12}mL^2
   \]

3. According to the problem, the rotational inertia is also given by \( md^2 \). Therefore, we can set these two expressions equal to each other:
   \[
   md^2 = \frac{1}{12}mL^2
   \]

4. To solve for \( d \), we divide both sides by \( m \):
   \[
   d^2 = \frac{1}{12}L^2
   \]

5. Taking the square root of both sides, we get:
   \[
   d = \frac{L}{2\sqrt{3}}
   \]

6. We need to find the ratio \( \frac{L}{d} \):
   \[
   \frac{L}{d} = \frac{L}{\frac{L}{2\sqrt{3}}} = 2\sqrt{3}
   \]

7. Therefore, the ratio \( \frac{L}{d} \) is \( 2\sqrt{3} \).

The final answer is \( \boxed{2\sqrt{3}} \).