find value of a so that the term independent of x in (rootx+a/x ssuare)10 is 405

(r+1)th term in the expansion of $(\sqrt{x}+\frac{a}{x^2}{)}^{10}$ is
$$\begin{gathered} {}^{10}C_r\left(\sqrt{x}{)}^{10-r}\right(\frac{a}{x^2}{)}^r \\ ={}^{10}C_r{a}^r*x^{5-\frac{r}{2}}*x^{-2r} \\ ={}^{10}C_r{a}^r*x^{5-\frac{r}{2}-2r} \\ ={}^{10}C_r{a}^r*x^{5-\frac{5r}{2}} \end{gathered}$$
let the term is independent of x , therefore exponent of x is zero
$$\begin{gathered} 5-\frac{5r}{2}=0 \\ \frac{r}{2}=1 \\ r=2 \end{gathered}$$
the coefficient of the term independent of x is 
$$\begin{gathered} {}^{10}C_2*a^2=405 \\ \frac{10*9}{2}{a}^2=405 \\ {a}^2=\frac{405}{45}=9 \\ a=\pm 3 \end{gathered}$$
thus the required value of a is 3 or -3.

hope this helps you