1. The problem asks for the average of the squares of the first 121 natural numbers. 2. The first 121 natural numbers are $1, 2, 3, \ldots, 121$. 3. The squares of these numbers are $1^2, 2^2, 3^2, \ldots, 121^2$. 4. We need to find the average of these squares, which is \[ \text{Average} = \frac{1^2 + 2^2 + 3^2 + \cdots + 121^2}{121}. \] 5. The formula for the sum of the squares of the first $n$ natural numbers is \[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. \] 6. Substituting $n=121$, we get \[ \sum_{k=1}^{121} k^2 = \frac{121 \times 122 \times 243}{6}. \] 7. Calculate the numerator: $121 \times 122 = 14762$, and $14762 \times 243 = 3586266$. 8. Divide by 6: \[ \frac{3586266}{6} = 597711. \] 9. Thus, \[ \sum_{k=1}^{121} k^2 = 597711. \] 10. Now find the average: \[ \text{Average} = \frac{597711}{121} = 4941. \] 11. Therefore, the average of the squares of the first 121 natural numbers is **4941**.