Coefficient

Answer is $960$ .

We have: $\displaystyle\left(2x^2-\dfrac{1}{x}\right)^{10}=\sum_{k=0}^{10}\binom{10}{k}\left(2x^2\right)^{10-k}.\left(-\dfrac{1}{x}\right)^k$

$\qquad\qquad\qquad\qquad\qquad\displaystyle=\sum_{k=0}^{10}\binom{10}{k}2^{10-k}(-1)^k.x^{20-3k}$

If $20-3k=-1$ , then $k=7$ .

So, the cofficient of $\dfrac{1}{x}$ is $\displaystyle\binom{10}{7}2^{10-7}(-1)^7=-960$ .

And the absolute value of it is $\boxed{960}$ .