Decomposition of cubes

The row corresponding to $n^3$ has $n$ numbers, beginning with $n^2-n+1$ and ending in $n^2+n-1$ . Therefore $n^2-n+1<2021, n^2+n-1>2021$ . Solving these two inequalities we get $n\leq 45, n\geq 45\implies n=\boxed {45}$ .

Visually

\[\begin{cases} 1^3 = 1 \\ 2^3 = 3 + 5 \\ 3^3 = 7 + 9 +11 \\ 4^3 = 13 + 15 + 17 + 19 \end{cases}

\quad \implies \quad

\begin{cases} 1^3 = \big(2({\color{Red}{1}}) - 1\big) \\ 2^3 = \big(2({\color{Red}{2}}) - 1\big) + \big(2({\color{Red}{3}}) - 1\big) \\ 3^3 = \big(2({\color{Red}{4}}) - 1\big) + \big(2({\color{Red}{5}}) - 1\big) + \big(2({\color{Red}{6}}) - 1\big) \\ 4^3 = \big(2({\color{Red}{7}}) - 1\big) + \big(2({\color{Red}{8}}) - 1\big) + \big(2({\color{Red}{9}}) - 1\big) + \big(2({\color{Red}{10}}) - 1\big) \end{cases}

\quad \implies \quad

\begin{cases} 1^3 = ({\color{Red}{1}}) \\ 2^3 = ({\color{Red}{2}}) + ({\color{Red}{3}}) \\ 3^3 = ({\color{Red}{4}}) + ({\color{Red}{5}}) + ({\color{Red}{6}}) \\ 4^3 = ({\color{Red}{7}}) + ({\color{Red}{8}}) + ({\color{Red}{9}}) + ({\color{Red}{10}}) \end{cases} \]

From this, $2021$ can be represented by $({\color{#D61F06}{1011}})$ can be. The row of $n^3$ contains numbers from $\frac{(n-1)(n)}{2}$ to $\frac{(n)(n+1)}{2}$