Given that ${\left(\frac{x}{4}-\frac{12}{x^2}\right)}^{12}$

$r^{th}$ term of an expression is $T_{r+1} ={}^{n}C_r{x}^{n-r}{y}^r$ $T_{r+1}={}^{12}C_r{\left(\frac{x}{4}\right)}^{12-r}{\left(-\frac{12}{x^2}\right)}^r={}^{12}C_r{\left(\frac{1}{4}\right)}^{12-r}(-12{)}^r·x^{12-r-2r}$
$\Rightarrow 12-3r=0$
$\Rightarrow r=4$

So term independent from $x$ is $T_5={}^{12}C_4·\frac{1}{4^8}·3^4·4^4={}^{12}C_4·\frac{3^4}{4^4}$

We know that ${}^{n}C_r=\frac{n!}{\left(n-r\right)!r!}$

$=55\times \frac{3^6}{4^4}$

They given that it is $=k \frac{3^6}{4^4}$

So required value of $k = 55.$