The coefficient of $$x^2$$ in the expansion of $$\left(2x^2 + \frac{1}{x}\right)^{10}$$, $$x \neq 0$$, is :

Expand using the Binomial Theorem:
$$(a+b)^{10}=\sum_{k=0}^{10} \binom{10}{k}\,a^{k}\,b^{\,10-k}$$
with $$a=2x^{2}$$ and $$b=\frac{1}{x}$$.

The general term is
$$T_k=\binom{10}{k}\,(2x^{2})^{k}\,\left(\frac{1}{x}\right)^{10-k}$$

Simplify the powers of $$x$$ in $$T_k$$:
$$(2x^{2})^{k}=2^{k}\,x^{2k}, \qquad \left(\frac{1}{x}\right)^{10-k}=x^{-(10-k)}$$
so
$$T_k=\binom{10}{k}\,2^{k}\,x^{2k}\,x^{-(10-k)} =\binom{10}{k}\,2^{k}\,x^{\,2k-(10-k)} =\binom{10}{k}\,2^{k}\,x^{\,3k-10}$$

To obtain the coefficient of $$x^{2}$$ set the exponent equal to 2:
$$3k-10 = 2 \;\Longrightarrow\; 3k = 12 \;\Longrightarrow\; k = 4$$

Substitute $$k=4$$ back into the coefficient part:
$$\binom{10}{4}\,2^{4} = 210 \times 16 = 3360$$

Hence, the coefficient of $$x^{2}$$ is $$3360$$.
Option B which is: $$3360$$