Two positive whole numbers differ by $3$. The sum of their squares is $117$. Find the larger of the two
numbers.
1. Let the two positive whole numbers be \( a \) and \( b \). Given that they differ by 3, we can write:
   \[
   a - b = 3
   \]
   Without loss of generality, assume \( a > b \). Therefore, we can write:
   \[
   a = b + 3
   \]

2. We are also given that the sum of their squares is 117:
   \[
   a^2 + b^2 = 117
   \]

3. Substitute \( a = b + 3 \) into the equation \( a^2 + b^2 = 117 \):
   \[
   (b + 3)^2 + b^2 = 117
   \]

4. Expand and simplify the equation:
   \[
   (b + 3)^2 + b^2 = 117
   \]
   \[
   b^2 + 6b + 9 + b^2 = 117
   \]
   \[
   2b^2 + 6b + 9 = 117
   \]

5. Subtract 117 from both sides to set the equation to zero:
   \[
   2b^2 + 6b + 9 - 117 = 0
   \]
   \[
   2b^2 + 6b - 108 = 0
   \]

6. Divide the entire equation by 2 to simplify:
   \[
   b^2 + 3b - 54 = 0
   \]

7. Solve the quadratic equation using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = 1 \), \( B = 3 \), and \( C = -54 \):
   \[
   b = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1}
   \]
   \[
   b = \frac{-3 \pm \sqrt{9 + 216}}{2}
   \]
   \[
   b = \frac{-3 \pm \sqrt{225}}{2}
   \]
   \[
   b = \frac{-3 \pm 15}{2}
   \]

8. This gives us two potential solutions for \( b \):
   \[
   b = \frac{-3 + 15}{2} = \frac{12}{2} = 6
   \]
   \[
   b = \frac{-3 - 15}{2} = \frac{-18}{2} = -9
   \]

9. Since \( b \) must be a positive whole number, we discard \( b = -9 \) and accept \( b = 6 \).

10. Substitute \( b = 6 \) back into \( a = b + 3 \) to find \( a \):
    \[
    a = 6 + 3 = 9
    \]

Conclusion:
\[
\boxed{9}
\]