Which of the following is a possible number of diagonals of a convex polygon?

(A) $02$  (B) $21$ (C) $32$ (D) $54$ (E) $63$
To determine the number of diagonals in a convex polygon with \( n \) sides, we use the formula:

\[
\text{Number of diagonals} = \frac{n(n-3)}{2}
\]

Let's verify each option by checking if it can be expressed in the form \(\frac{n(n-3)}{2}\) for some integer \( n \).

1. **Option (A): 2**

   \[
   \frac{n(n-3)}{2} = 2 \implies n(n-3) = 4
   \]

   Solving the quadratic equation:

   \[
   n^2 - 3n - 4 = 0
   \]

   Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

   \[
   n = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}
   \]

   This gives us:

   \[
   n = 4 \quad \text{or} \quad n = -1
   \]

   Since \( n \) must be a positive integer, \( n = 4 \). For \( n = 4 \) (a quadrilateral), the number of diagonals is:

   \[
   \frac{4(4-3)}{2} = 2
   \]

   So, option (A) is possible.

2. **Option (B): 21**

   \[
   \frac{n(n-3)}{2} = 21 \implies n(n-3) = 42
   \]

   Solving the quadratic equation:

   \[
   n^2 - 3n - 42 = 0
   \]

   Using the quadratic formula:

   \[
   n = \frac{3 \pm \sqrt{9 + 168}}{2} = \frac{3 \pm 13}{2}
   \]

   This gives us:

   \[
   n = 8 \quad \text{or} \quad n = -5
   \]

   Since \( n \) must be a positive integer, \( n = 8 \). For \( n = 8 \) (an octagon), the number of diagonals is:

   \[
   \frac{8(8-3)}{2} = 20
   \]

   So, option (B) is not possible.

3. **Option (C): 32**

   \[
   \frac{n(n-3)}{2} = 32 \implies n(n-3) = 64
   \]

   Solving the quadratic equation:

   \[
   n^2 - 3n - 64 = 0
   \]

   Using the quadratic formula:

   \[
   n = \frac{3 \pm \sqrt{9 + 256}}{2} = \frac{3 \pm 17}{2}
   \]

   This gives us:

   \[
   n = 10 \quad \text{or} \quad n = -7
   \]

   Since \( n \) must be a positive integer, \( n = 10 \). For \( n = 10 \) (a decagon), the number of diagonals is:

   \[
   \frac{10(10-3)}{2} = 35
   \]

   So, option (C) is not possible.

4. **Option (D): 54**

   \[
   \frac{n(n-3)}{2} = 54 \implies n(n-3) = 108
   \]

   Solving the quadratic equation:

   \[
   n^2 - 3n - 108 = 0
   \]

   Using the quadratic formula:

   \[
   n = \frac{3 \pm \sqrt{9 + 432}}{2} = \frac{3 \pm 21}{2}
   \]

   This gives us:

   \[
   n = 12 \quad \text{or} \quad n = -9
   \]

   Since \( n \) must be a positive integer, \( n = 12 \). For \( n = 12 \) (a dodecagon), the number of diagonals is:

   \[
   \frac{12(12-3)}{2} = 54
   \]

   So, option (D) is possible.

5. **Option (E): 63**

   \[
   \frac{n(n-3)}{2} = 63 \implies n(n-3) = 126
   \]

   Solving the quadratic equation:

   \[
   n^2 - 3n - 126 = 0
   \]

   Using the quadratic formula:

   \[
   n = \frac{3 \pm \sqrt{9 + 504}}{2} = \frac{3 \pm 23}{2}
   \]

   This gives us:

   \[
   n = 13 \quad \text{or} \quad n = -10
   \]

   Since \( n \) must be a positive integer, \( n = 13 \). For \( n = 13 \) (a tridecagon), the number of diagonals is:

   \[
   \frac{13(13-3)}{2} = 65
   \]

   So, option (E) is not possible.

The final answer is \(\boxed{02 \text{ and } 54}\)