Let $$f$$ be a polynomial function such that $$f(3x) = f'(x) \cdot f''(x)$$, for all $$x \in R$$. Then:

Assume that the degree of the required polynomial is $$n$$. The left-hand expression $$f(3x)$$ clearly has degree $$n$$, while the right-hand expression $$f'(x)\cdot f''(x)$$ has degree $$(n-1)+(n-2)=2n-3$$. Equality for every real $$x$$ forces the two degrees to match, so

$$n = 2n-3 \;\;\Longrightarrow\;\; n=3.$$

Hence $$f$$ must be a cubic. Write it in its most general form

$$f(x)=ax^{3}+bx^{2}+cx+d.\quad -(1)$$

Because $$a\neq 0$$ (otherwise the degree would drop), calculate the first two derivatives

$$f'(x)=3ax^{2}+2bx+c,\quad -(2)$$

$$f''(x)=6ax+2b.\quad -(3)$$

Form the product demanded on the right-hand side of the functional equation:

$$\begin{aligned} f'(x)\,f''(x) &=(3ax^{2}+2bx+c)(6ax+2b) \\ &=3ax^{2}\cdot6ax + 3ax^{2}\cdot2b + 2bx\cdot6ax + 2bx\cdot2b + c\cdot6ax + c\cdot2b \\ &=18a^{2}x^{3} + 6abx^{2} + 12abx^{2} + 4b^{2}x + 6acx + 2bc \\ &=18a^{2}x^{3} + 18abx^{2} + (4b^{2}+6ac)x + 2bc. \quad -(4) \end{aligned}$$

Now expand the left-hand side $$f(3x)$$:

$$f(3x)=a(3x)^{3}+b(3x)^{2}+c(3x)+d =27ax^{3}+9bx^{2}+3cx+d.\quad -(5)$$

Since $$f(3x) = f'(x)\,f''(x)$$ for every $$x$$, equate the coefficients of corresponding powers of $$x$$ in (4) and (5).

Cubic term, $$x^{3}$$:

$$18a^{2}=27a \;\;\Longrightarrow\;\; 18a^{2}-27a=0 \;\;\Longrightarrow\;\; 9a(2a-3)=0 \;\;\Longrightarrow\;\; a=\dfrac32.\quad -(6)$$

Quadratic term, $$x^{2}$$:

$$18ab=9b \;\;\Longrightarrow\;\; 18ab-9b=0 \;\;\Longrightarrow\;\; 9b(2a-1)=0.$$

With $$a=\dfrac32$$ we have $$2a-1=2\neq0,$$ hence

$$b=0.\quad -(7)$$

Linear term, $$x^{1}$$:

$$4b^{2}+6ac=3c.$$

Insert $$b=0$$ to obtain $$6ac=3c,$$ that is

$$3c(2a-1)=0.$$

Again $$2a-1=2\neq0,$$ so

$$c=0.\quad -(8)$$

Constant term:

The constant on the right in (4) is $$2bc$$, which vanishes because $$b=0$$ or $$c=0$$ (in fact both). Hence

$$d=0.\quad -(9)$$

All coefficients are now fixed; the unique polynomial satisfying the given condition is therefore

$$f(x)=\dfrac32\,x^{3}.\quad -(10)$$

For the values asked in the options, compute

$$f(2)=\dfrac32\,(2)^{3}=\dfrac32\cdot8=12,\quad -(11)$$

$$f'(x)=\dfrac{9}{2}x^{2}\;\Longrightarrow\; f'(2)=\dfrac{9}{2}\,(2)^{2}=\dfrac{9}{2}\cdot4=18,\quad -(12)$$

$$f''(x)=9x\;\Longrightarrow\; f''(2)=9\cdot2=18.\quad -(13)$$

Check each option:

Option A: $$f(2)+f'(2)=12+18=30\neq28.$$

Option B: $$f''(2)-f'(2)=18-18=0,$$ which is true.

Option C: $$f(2)-f'(2)+f''(2)=12-18+18=12\neq10.$$

Option D: $$f''(2)-f(2)=18-12=6\neq4.$$

Only Option B is satisfied.

Hence, the correct answer is Option B.