If the constant term in the binomial expansion of ${\left(\sqrt{\mathrm{x}}-\frac{\mathrm{k}}{{\mathrm{x}}^2}\right)}^{10}$ is 405, then $\left|\mathrm{k}\right|=$

In ${\left(\sqrt{\mathrm{x}}-\frac{\mathrm{k}}{{\mathrm{x}}^2}\right)}^{10};{\mathrm{T}}_{\mathrm{r}+1}{=}^{10}{\mathrm{C}}_{\mathrm{r}}\left(\sqrt{\mathrm{x}}{)}^{10-\mathrm{r}}{\left(-\frac{\mathrm{k}}{{\mathrm{x}}^2}\right)}^{\mathrm{r}}{=}^{10}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{(10-\mathrm{r})/2-2\mathrm{r}}\right(-\mathrm{k}{)}^{\mathrm{r}}$

$\frac{10-\mathrm{r}}{2}-2\mathrm{r}=0\Rightarrow 10-5\mathrm{r}=0\Rightarrow \mathrm{r}=2$

$\therefore$ Constant term is ${ }^{10}{\mathrm{C}}_2(-\mathrm{k}{)}^2=45{\mathrm{k}}^2\Rightarrow 405=45{\mathrm{k}}^2\Rightarrow {\mathrm{k}}^2=9\Rightarrow |\mathrm{k}|=3$.