Bit of an overkill, but an alternative approach giving a bound on the rate of convergence as well:
$$
S_{n}:=\sum_{i=1}^{3n^{2}}\frac{1}{\sqrt[3]{\left(8n^{3}+2i\right)^{2}}}=\frac{1}{4n^{2}}\sum_{i=1}^{3n^{2}}\frac{1}{\sqrt[3]{\left(1+\frac{2i}{8n^{3}}\right)^{2}}}
$$
By Taylor expansion, for small enough $x$:
$$
1-\frac{2}{3}x\leq\frac{1}{\left(1+x\right)^{2/3}}\leq1-\frac{2}{3}x+\frac{5}{9}x^{2}
$$
Some double-checking if this really holds:
$$
1-\left(1+x\right)^{2}\left(1-\frac{2}{3}x\right)^{3}=\frac{1}{27}x^{2}\left(8x^{3}-20x^{2}-10x+45\right)
$$
This is positive for $1\geq x\geq0$ as $8x^{3}-20x^{2}-10x+45\geq45-30\geq15$.
$$
1-\left(1+x\right)^{2}\left(1-\frac{2}{3}x+\frac{5}{9}x^{2}\right)^{3}=-\frac{1}{729}x^{3}\left(125x^{5}-200x^{4}+440x^{3}+144x^{2}-270x+1080\right)
$$
This is negative for $1\geq x\geq0$, as $125x^{5}-200x^{4}+440x^{3}+144x^{2}-270x+1080\geq-200-270+1080>0$.
As $6n^2<8n^3$ (maximum of $i$ for a given $n$), these estimations hold for every $n$.
Using these approximations to bound the sum:
$$
S_{n,0}:=\frac{1}{4n^{2}}\cdot\sum_{i=1}^{3n^{2}}1=\frac{3}{4}
$$
$$
S_{n,1}:=\frac{1}{4n^{2}}\cdot\frac{2}{3}\sum_{i=1}^{3n^{2}}\frac{2i}{8n^{3}}=\frac{1}{4n^{2}}\cdot\frac{2}{3}\cdot\frac{3n^{2}+1}{4n}= \mathcal{O}\left( \frac{1}{n}\right)
$$
$$
S_{n,2}:=\frac{1}{4n^{2}}\cdot\frac{5}{9}\sum_{i=1}^{3n^{2}}\left(\frac{2i}{8n^{3}}\right)^{2}=\frac{1}{4n^{2}}\cdot\frac{5}{9}\frac{\left(3n^{2}+1\right)\left(6n^{2}+1\right)}{48n^{4}}= \mathcal{O}\left( \frac{1}{n^2}\right)
$$
Hence:
$$
\frac{3}{4}+S_{n,1}+S_{n,2}\geq S_{n}\geq\frac{3}{4}+S_{n,0}
$$
And the rate of the convergence is $\mathcal{O}\left( \frac{1}{n}\right)$
