The polynomial $ x^3\minus{}ax^2\plus{}bx\minus{}2010$ has three positive integer zeros. What is the smallest possible value of $ a$?

$ \textbf{(A)}\ 78 \qquad
\textbf{(B)}\ 88 \qquad
\textbf{(C)}\ 98 \qquad
\textbf{(D)}\ 108 \qquad
\textbf{(E)}\ 118$
1. Let the polynomial be \( P(x) = x^3 - ax^2 + bx - 2010 \). Given that it has three positive integer zeros, let these zeros be \( \alpha, \beta, \gamma \).

2. By Vieta's formulas, we know:
   \[
   \alpha + \beta + \gamma = a
   \]
   \[
   \alpha \beta + \beta \gamma + \gamma \alpha = b
   \]
   \[
   \alpha \beta \gamma = 2010
   \]

3. We need to find the smallest possible value of \( a \). First, factorize 2010 into its prime factors:
   \[
   2010 = 2 \times 3 \times 5 \times 67
   \]

4. To minimize \( a = \alpha + \beta + \gamma \), we should choose the factors such that their sum is minimized. The factors of 2010 are \( 2, 3, 5, \) and \( 67 \).

5. We can combine these factors into three groups to minimize their sum. The best way to do this is:
   \[
   \alpha = 2 \times 3 = 6, \quad \beta = 5, \quad \gamma = 67
   \]

6. Now, calculate \( a \):
   \[
   a = \alpha + \beta + \gamma = 6 + 5 + 67 = 78
   \]

7. Verify that these values satisfy the conditions:
   \[
   \alpha \beta \gamma = 6 \times 5 \times 67 = 2010
   \]

Thus, the smallest possible value of \( a \) is \( 78 \).

The final answer is \( \boxed{78} \)