$|3x — 2|$ is an absolute value of 3x-2
Now, whenever there is an absolute value we need to consider two cases

Case 1:Whatever is inside the modulus is >= 0 3x-2 >= 0
If the value inside the modulus is non-negative then the modulus can be removed directly

As 3x-2 >= 0 which means 3x >=2 or $x >= \frac{2}{3}$

so $|3x — 2|$ = 3x - 2
=> $3x — 2 < 8$
=> 3x < 8+2 => 3x < 10
=> x < $\frac{10}{3}$
Now we have two conditions for x
$\frac{2}{3} <= x < \frac{10}{3}$
So, x = 1,2,3

Case 2: Whatever is inside the modulus is < 0 3x-2 < 0
If the value inside the modulus is negative then the modulus can be removed after multiplying the terms inside the modulus by -1

As 3x-2 >= 0 which means 3x < 2 or $x < \frac{2}{3}$
so $|3x — 2|$ = -(3x - 2)
=> $-(3x — 2) < 8$
=> 3x > 2 - 8 => 3x > -6
=> x > $\frac{-6}{3}$
=> x > -2
Now we have two conditions for x
$-2 < x < \frac{2}{3}$
So, x = -1,0

So, there are 5 values of x which are possible
x = -1,0,1,2,3

So, Answer will be E
Hope it helps!

Watch the following video to learn how to Solve Inequality + Absolute value Problems