$$\begin{gathered} \text{Let}f(x)=\underset{n\to \infty }{\lim }\left(\frac{1}{n^3}\sum _{k=1}^n\right[\frac{k^2}{3^x}\left]\right)\text{, where [ ] denotes the greatest integer function, then}12\sum _{j=1}^{\mathrm{\infty }}f\mathrm{(\; j\; )}\text{ is equal to} \\ \end{gathered}$$

The correct answer is 2.

To find the value of the given expression, we first need to evaluate the function f(x) using the properties of the greatest integer function and the Squeeze Theorem (Sandwich Theorem).

Step 1: Apply bounds for the greatest integer function
For any real number y, the greatest integer function \left[y\right] satisfies the fundamental inequality:
y-1<\left[y\right]\le y
Substituting y=\frac{k^2}{3^x} into the inequality, we get:
\frac{k^2}{3^x}-1<\left[\frac{k^2}{3^x}\right]\le \frac{k^2}{3^x}

Step 2: Sum the inequality from k = 1 to n
Summing all terms from k=1 to n:
\sum _{k=1}^n(\frac{k^2}{3^x}-1)<\sum _{k=1}^n\left[\frac{k^2}{3^x}\right]\le \sum _{k=1}^n\frac{k^2}{3^x}
Factoring out terms independent of k:
\frac{1}{3^x}\sum _{k=1}^n{k}^2-n<\sum _{k=1}^n\left[\frac{k^2}{3^x}\right]\le \frac{1}{3^x}\sum _{k=1}^n{k}^2

Step 3: Divide by n³ and apply the limit as n → ∞
Multiplying the entire inequality by \frac{1}{n^3} gives:
\frac{1}{3^x}·\frac{1}{n^3}\sum _{k=1}^n{k}^2-\frac{1}{n^2}<\frac{1}{n^3}\sum _{k=1}^n\left[\frac{k^2}{3^x}\right]\le \frac{1}{3^x}·\frac{1}{n^3}\sum _{k=1}^n{k}^2
Using the standard summation formula for the sum of squares of first n natural numbers:
\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}
We evaluate the limiting value:
\underset{n\to \infty }{\lim }\frac{1}{n^3}\sum _{k=1}^n{k}^2=\underset{n\to \infty }{\lim }\frac{n(n+1)(2n+1)}{6n^3}=\frac{2}{6}=\frac{1}{3}

Now, taking the limit as n\to \infty on both ends of the inequality:
Lower bound limit:
\underset{n\to \infty }{\lim }(\frac{1}{3^x}·\frac{1}{n^3}\sum _{k=1}^n{k}^2-\frac{1}{n^2})=\frac{1}{3^x}·\frac{1}{3}-0=\frac{1}{3^{x+1}}
Upper bound limit:
\underset{n\to \infty }{\lim }(\frac{1}{3^x}·\frac{1}{n^3}\sum _{k=1}^n{k}^2)=\frac{1}{3^x}·\frac{1}{3}=\frac{1}{3^{x+1}}
By the Squeeze Theorem, the middle limit f(x) is equal to:
f(x)=\frac{1}{3^{x+1}}

Step 4: Compute the given infinite sum
We are asked to evaluate:
12\sum _{j=1}^{\infty }f(j)=12\sum _{j=1}^{\infty }\frac{1}{3^{j+1}}
Expanding the infinite series:
\sum _{j=1}^{\infty }\frac{1}{3^{j+1}}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\dots
This is an infinite geometric series with first term a=\frac{1}{9} and common ratio r=\frac{1}{3}.
Using the formula for the sum of an infinite geometric series S=\frac{a}{1-r}:
\sum _{j=1}^{\infty }\frac{1}{3^{j+1}}=\frac{\frac{1}{9}}{1-\frac{1}{3}}=\frac{\frac{1}{9}}{\frac{2}{3}}=\frac{1}{6}

Multiplying by 12:
12\sum _{j=1}^{\infty }f(j)=12·\frac{1}{6}=2

Thus, the final answer is 2.