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

First recall the binomial theorem: for any two terms $$a$$ and $$b$$ and a positive integer $$n$$, the expansion is $$ (a+b)^n=\displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{\,n-r}b^{\,r}\,. $$

Here we have $$a=\sqrt{x}=x^{1/2},\; b=-\dfrac{k}{x^{2}}= -k\,x^{-2},\; n=10.$$

So the general term in the expansion is obtained by choosing the $$r^{\text{th}}$$ term (starting from $$r=0$$):

$$ T_r=\binom{10}{r}\left(\sqrt{x}\right)^{\,10-r}\!\left(-\dfrac{k}{x^{2}}\right)^{r}. $$

We simplify the powers of $$x$$ inside $$T_r$$. Using the laws of indices $$x^{m}\,x^{n}=x^{m+n}$$ and $$(x^{m})^{n}=x^{mn},$$ we write

$$\left(\sqrt{x}\right)^{\,10-r}=x^{\frac{1}{2}(10-r)} \quad\text{and}\quad \left(-\dfrac{k}{x^{2}}\right)^{r}=(-k)^{r}\,x^{-2r}.$$ Therefore

$$ T_r=\binom{10}{r}(-k)^{r}\,x^{\frac{1}{2}(10-r)}\,x^{-2r} =\binom{10}{r}(-k)^{r}\,x^{\,\frac{10-r}{2}-2r}. $$

The exponent of $$x$$ in $$T_r$$ is

$$ \frac{10-r}{2}-2r =5-\frac{r}{2}-2r =5-\frac{5r}{2}. $$

For the term to be constant (i.e., independent of $$x$$) this exponent must be zero:

$$ 5-\frac{5r}{2}=0 \;\Longrightarrow\; \frac{5r}{2}=5 \;\Longrightarrow\; r=2. $$

Thus the constant term comes from $$r=2$$. We now substitute $$r=2$$ back into the expression for $$T_r$$ (noting that $$(-k)^2=k^2$$ because the square removes the minus sign):

$$ T_2=\binom{10}{2}\,k^{2}\,x^{\,\frac{10-2}{2}-2\!\cdot\!2} =\binom{10}{2}\,k^{2}\,x^{\,4-4} =\binom{10}{2}\,k^{2}\,x^{0} =\binom{10}{2}\,k^{2}. $$

Since $$\displaystyle\binom{10}{2}=\dfrac{10\cdot9}{2}=45,$$ we have

$$ T_2 = 45\,k^{2}. $$

The problem states that this constant term equals 405, so

$$ 45\,k^{2}=405 \;\Longrightarrow\; k^{2}= \frac{405}{45}=9. $$

Taking square roots, $$|k|=\sqrt{9}=3.$$

Hence, the correct answer is Option C.