1. Consider the binomial expansion $\left(\frac{x}{4}-\frac{12}{x^2}\right)^{12}.$

The general term is $T_{r+1}=\binom{12}{r}\left(\frac{x}{4}\right)^{12-r}\left(-\frac{12}{x^2}\right)^r.$

2. Simplify the general term: $T_{r+1}=\binom{12}{r}\frac{x^{12-r}}{4^{12-r}}\cdot (-12)^r x^{-2r}$ $=\binom{12}{r}\frac{(-12)^r}{4^{12-r}}x^{12-r-2r}$ $=\binom{12}{r}\frac{(-12)^r}{4^{12-r}}x^{12-3r}.$

3. For the term independent of $x$, the power of $x$ must be zero: $12-3r=0$ $r=4.$

So the constant term is $T_5$.

4. Compute $T_5$: $T_5=\binom{12}{4}\left(\frac{x}{4}\right)^8\left(-\frac{12}{x^2}\right)^4.$

Since $(-12)^4=12^4$, $T_5=\binom{12}{4}\frac{12^4}{4^8}.$

Now, $\binom{12}{4}=495,$ and $12^4=(3\cdot 4)^4=3^4\cdot 4^4.$

Hence, $T_5=495\cdot \frac{3^4\cdot 4^4}{4^8}=495\cdot \frac{3^4}{4^4}.$

5. The constant term is given to be $\left(\frac{3^6}{4^4}\right)k.$

So equate: $\left(\frac{3^6}{4^4}\right)k = 495\cdot \frac{3^4}{4^4}.$

Multiply both sides by $\frac{4^4}{3^4}$: $3^2 k = 495$ $9k=495$ $k=55.$

6. Therefore, $\boxed{55}.$