$\lim _{n \rightarrow \infty} \frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\ldots \ldots \ldots \ldots+\frac{n^2}{n^3+1}=$

To evaluate the limit, we express the sum using summation notation:

$\lim_{n \to \infty} \frac{1}{n^3+1} (1^2 + 2^2 + 3^2 + \cdots + n^2)$.

The sum of the first $n$ squares is $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, which gives:

$\lim_{n \to \infty} \frac{1}{n^3+1} \cdot \frac{n(n+1)(2n+1)}{6} = \lim_{n \to \infty} \frac{2n^3 + 3n^2 + n}{6n^3 + 6}$.

Dividing numerator and denominator by $n^3$:

$\lim_{n \to \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6 + \frac{6}{n^3}} = \frac{2 + 0 + 0}{6 + 0} = \frac{1}{3}$.

The limit is $\frac{1}{3}$, so the answer is B.