If the  term independent of x  in the binomial expansion ${(\sqrt{x}-\frac{k}{x^2})}^{10}$  is 405 then the value of $k^2$  is  

$\begin{array}{r}{\left(\sqrt{x}+\frac{K}{x^2}\right)}^{10}\\ T_{r+1}{=}^{10}{C}_r(\sqrt{x}{)}^{10-r}{\left(\frac{K}{x^2}\right)}^r\\ {=}^{10}{C}_r(x{)}^{\frac{10-r}{2}}{K}^r{x}^{-2r}\end{array}$

For term independent of x 

$$\begin{gathered} \text{ So }\frac{10-r}{2}-2r=0 \\ \Rightarrow 10-5r=0\Rightarrow r=\frac{10}{5}=2 \end{gathered}$$

$\begin{array}{r}{ }^{10}{\mathrm{C}}_2{\mathrm{K}}^2=\frac{10\times 9}{1\times 2}\times K^2=405\\ \Rightarrow K^2=9\end{array}$