A certain family has 3 sons: Richard is 6 years older than

Let Richard's age = r, David's age = d, and Scott's age = s.
$$r=d+6$$
$$d=s+8$$
In 8 years, 'r' will be twice as old as Scott
$$r+8=2\left(s+8\right)$$
$$r+8=2s+16$$
$$r=2s+8$$
$$Recall\ that,\ r=d+6,\ where\ d=s+8$$
$$r=s+8+6=s+14$$
$$where,\ r=2s+8$$
$$2s+8=s+14$$
$$s=6$$
$$r=20$$
$$and\ d=14$$
$$David's\ age\ 4\ years\ ago=>d-4$$
$$=14-4=10\ years\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(Answer=Option\ B\right)$$

\(R\) is \(6\) years older than \(D\), \(D\) is \(8\) years older than \(S\)

So \(R\) is \((6+8)=14\) years older than \(S\), so we can say that \(R = S + 14\)

After \(8\) years, age of Scott \(= S+8\), age of Richard \(= S+8+14 = S+22\) (the difference will always be \(14\) years as long as both are alive)
Given: \(S+22 = 2 \cdot (S+8)\)
Solve to get \(S=6\)

Thus currently \(S\) is \(6\) years, \(4\) years ago \(S\) would have been \(2\) years old, and David must have been \(2+8 = 10\) years old (the difference between their ages would have been same then as now)

Hence, the correct answer is B