Question:

The general term in the binomial expansion of $\left(\sqrt{x} - \frac{k}{x}\right)^{10}$ is given by $$T_{r+1} = \binom{10}{r} (\sqrt{x})^{10-r} \left(-\frac{k}{x}\right)^r$$ $$T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (-k)^r (x^{-1})^r$$ $$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-r}{2}} x^{-r}$$ $$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-r}{2} - r}$$ $$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-3r}{2}}$$ For the term to be constant, the power of $x$ must be zero. $$\frac{10-3r}{2} = 0$$ $$10-3r = 0$$ $$3r = 10$$ $$r = \frac{10}{3}$$ Since $r$ must be an integer, there is no constant term in the expansion. Let's re-evaluate the problem statement. It is possible that the problem intended for the expression to be $\left(\sqrt{x} - \frac{k}{x^2}\right)^{10}$ or similar, or that the question is flawed. However, assuming the question is valid and there is a constant term, there must be a misinterpretation of the problem or a typo in the provided solution. Let's assume the problem meant $\left(\sqrt{x} - \frac{k}{x^2}\right)^{10}$. Then the general term would be: $$T_{r+1} = \binom{10}{r} (\sqrt{x})^{10-r} \left(-\frac{k}{x^2}\right)^r$$ $$T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (-k)^r (x^{-2})^r$$ $$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-r}{2}} x^{-2r}$$ $$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-r}{2} - 2r}$$ $$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-r-4r}{2}}$$ $$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-5r}{2}}$$ For the term to be constant, the power of $x$ must be zero. $$\frac{10-5r}{2} = 0$$ $$10-5r = 0$$ $$5r = 10$$ $$r = 2$$ This is an integer, so there is a constant term when $r=2$. The constant term is $T_{2+1} = T_3$. $$T_3 = \binom{10}{2} (-k)^2$$ $$T_3 = \frac{10 \times 9}{2} k^2$$ $$T_3 = 45k^2$$ Given that the constant term is 405: $$45k^2 = 405$$ $$k^2 = \frac{405}{45}$$ $$k^2 = 9$$ $$k = \pm 3$$ Therefore, $|k| = 3$. The final answer is $\boxed{3}$.