A function $ f$ from the integers to the integers is defined as follows:
\[ f(n) \equal{} \begin{cases} n \plus{} 3 & \text{if n is odd} \\
n/2 & \text{if n is even} \end{cases}
\]Suppose $ k$ is odd and $ f(f(f(k))) \equal{} 27$. What is the sum of the digits of $ k$?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$
1. Given the function \( f \) defined as:
   \[
   f(n) = \begin{cases} 
   n + 3 & \text{if } n \text{ is odd} \\
   \frac{n}{2} & \text{if } n \text{ is even} 
   \end{cases}
   \]
   and the condition \( f(f(f(k))) = 27 \) with \( k \) being odd.

2. Since \( k \) is odd, we start by applying the function \( f \) to \( k \):
   \[
   f(k) = k + 3
   \]
   Since \( k \) is odd, \( k + 3 \) is even.

3. Next, we apply the function \( f \) to \( k + 3 \):
   \[
   f(k + 3) = \frac{k + 3}{2}
   \]
   Since \( k + 3 \) is even, \( \frac{k + 3}{2} \) is an integer.

4. Finally, we apply the function \( f \) to \( \frac{k + 3}{2} \):
   \[
   f\left(\frac{k + 3}{2}\right) = \begin{cases} 
   \frac{k + 3}{2} + 3 & \text{if } \frac{k + 3}{2} \text{ is odd} \\
   \frac{\frac{k + 3}{2}}{2} & \text{if } \frac{k + 3}{2} \text{ is even} 
   \end{cases}
   \]
   Since we need \( f(f(f(k))) = 27 \), we set:
   \[
   f\left(\frac{k + 3}{2}\right) = 27
   \]

5. We need to determine whether \( \frac{k + 3}{2} \) is odd or even. Let's assume \( \frac{k + 3}{2} \) is even:
   \[
   f\left(\frac{k + 3}{2}\right) = \frac{\frac{k + 3}{2}}{2} = \frac{k + 3}{4}
   \]
   Setting this equal to 27:
   \[
   \frac{k + 3}{4} = 27
   \]
   Solving for \( k \):
   \[
   k + 3 = 108 \implies k = 105
   \]

6. Verify the solution:
   \[
   f(105) = 105 + 3 = 108
   \]
   \[
   f(108) = \frac{108}{2} = 54
   \]
   \[
   f(54) = \frac{54}{2} = 27
   \]
   This confirms \( f(f(f(105))) = 27 \).

7. The sum of the digits of \( k = 105 \) is:
   \[
   1 + 0 + 5 = 6
   \]

The final answer is \(\boxed{6}\).