Let the present age of the husband be x years and the present age of his wife be y years.
Since, the husband is 6 years older than his wife, we can write the equation as:
x = y + 6 ............................(1)

We are also given that 20 years ago, the husband was 15 times older than his son, which implies that the husband’s age 20 years ago was 15 times the age of the son 20 years ago.
Let the age of the son 20 years ago be z years.
Therefore, the husband’s age 20 years ago = 15z ............................(2)

Also, we are given that the wife is twice as old as the son.
Therefore, the wife’s age at present = 2z ...............................(3)

Now, we have three equations (1), (2) and (3) with three unknowns, x, y and z.
Solving them, we get:
x = 15z + 6
y = 2z
Substituting the value of y in equation (1), we get
x = 2z + 6
Therefore, substituting the value of x in equation (2), we get
15z + 6 = 15z
⇒ 6 = 0

This is impossible. Hence, the given statement is false and the given options are not possible.

Therefore, the correct answer is Option B (44 years).

If you think the solution is wrong then please provide your own solution below in the comments section .