= Solution
Let $S_N=N^{-3}\sum_{n=1}^N n^2$. The expression in part (ii) gives
$$
T_N=3\left(S_N-\frac1N\right)+\frac3{2N}=3S_N-\frac3{2N}.
$$
Because $3x^2$ is <continuous> on $[0,1]$, the <composite trapezoidal rule> converges to its <definite integral>, which is $1$ by part (i). In fact its standard <composite trapezoidal error bound> is $|T_N-1|\leq1/(2N^2)$, since $f''=6$. Therefore
$$
\boxed{\lim_{N\to\infty}S_N=\lim_{N\to\infty}\left(\frac{T_N}{3}+\frac1{2N}\right)=\frac13.}
$$
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