If the term independent of $x$ in the $\left( \sqrt{x} - \frac{k}{x^2} \right)^{10}$ is 405, then $k$ equals

Using the general term, the $(r+1)$th term is $$T_{r+1}=\binom{10}{r}(\sqrt{x})^{10-r}\left(-\frac{k}{x^2}\right)^r.$$ The power of $x$ in this term is $$\frac{10-r}{2}-2r=5-\frac{5r}{2}.$$ For the term independent of $x$, set the exponent to $0$: $$5-\frac{5r}{2}=0.$$ So $$r=2.$$ Hence the constant term is $$\binom{10}{2}(\sqrt{x})^8\left(-\frac{k}{x^2}\right)^2=45k^2.$$ Given this is $405$, $$45k^2=405.$$ Thus $$k^2=9.$$ Therefore $$k=\pm 3.$$ Matching with the options, this is $3,-3$, so the correct choice is $B$.