1. We are given the quadratic inequality

$x^2 - 2(3k-1)x + 8k^2 - 7 > 0$

which must be valid for every $x \in \mathbb{R}$.

2. For a quadratic

$ax^2+bx+c>0 \quad \text{for all } x\in\mathbb{R},$

the necessary and sufficient conditions are:

• $a>0$
• discriminant $\Delta<0$

Here,

$a=1>0$

so we only need:

$\Delta<0.$

3. Identify coefficients:

$a=1,\quad b=-2(3k-1),\quad c=8k^2-7.$

Thus,

$\Delta=b^2-4ac=[-2(3k-1)]^2-4(1)(8k^2-7).$

4. Simplify:

$[-2(3k-1)]^2=4(3k-1)^2$

So,

$\Delta=4(3k-1)^2-4(8k^2-7)$

$=4\big[(3k-1)^2-(8k^2-7)\big].$

Now,

$(3k-1)^2=9k^2-6k+1$

Hence,

$\Delta=4\big[(9k^2-6k+1)-8k^2+7\big]$

$=4(k^2-6k+8)$

$=4(k-2)(k-4).$

5. For positivity for all real $x$:

$\Delta<0 \Rightarrow 4(k-2)(k-4)<0$

$\Rightarrow (k-2)(k-4)<0.$

This holds when

$2<k<4.$

6. Since $k$ is an integer, the only possible value is

$k=3.$

7. Check options:

• A: $4$ → gives $\Delta=0$, not strictly positive for all $x$
• B: $2$ → gives $\Delta=0$, not strictly positive for all $x$
• C: $3$ → valid
• D: $0$ → gives $\Delta>0$, not valid

Therefore, the correct option is:

$\boxed{3}$