Use the squeeze inequality for the floor: $\frac{k^2}{3^x}-1<\left[\frac{k^2}{3^x}\right]\le \frac{k^2}{3^x}$. Summing for $k=1$ to $n$ and dividing by $n^3$:

$\frac{1}{n^3}\left(\frac{1}{3^x}\sum k^2\right)-\frac{1}{n^2}<\frac{1}{n^3}\sum \left[\frac{k^2}{3^x}\right]\le \frac{1}{n^3\cdot 3^x}\sum k^2$.

Now ${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$, so both bounds tend to $\frac{1}{3\cdot 3^x}=\frac{1}{3^{x+1}}$ as $n\to \infty$. Thus $f(x)=\frac{1}{3^{x+1}}$.

Sum the geometric series: ${\sum }_{j=1}^{\infty }f(j)={\sum }_{j=1}^{\infty }\frac{1}{3^{j+1}}=\frac{1}{9}\cdot \frac{1}{1-1/3}=\frac{1}{9}\cdot \frac{3}{2}=\frac{1}{6}$.

Multiply by 12: $12\cdot \frac{1}{6}=2$.