1. **State the problem:** Evaluate the limit as $n$ approaches infinity of the expression $$6 \frac{1^2 + 2^2 + 3^2 + \cdots + n^2}{n^3}.$$\n\n2. **Recall the formula for the sum of squares:** The sum of the first $n$ squares is given by $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}.$$\n\n3. **Substitute the sum into the expression:**\n$$6 \frac{\frac{n(n+1)(2n+1)}{6}}{n^3} = 6 \times \frac{n(n+1)(2n+1)}{6 n^3} = \frac{n(n+1)(2n+1)}{n^3}.$$\n\n4. **Simplify the expression:**\n$$\frac{n(n+1)(2n+1)}{n^3} = \frac{(n+1)(2n+1)}{n^2} = \frac{(n+1)}{n} \times \frac{(2n+1)}{n}.$$\n\n5. **Rewrite each factor:**\n$$\frac{n+1}{n} = 1 + \frac{1}{n}, \quad \frac{2n+1}{n} = 2 + \frac{1}{n}.$$\n\n6. **Take the limit as $n \to \infty$:**\n$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right) \left(2 + \frac{1}{n}\right) = (1 + 0)(2 + 0) = 1 \times 2 = 2.$$\n\n**Final answer:** $$\boxed{2}.$$