Let the man's speed be $v_m$ and the son's speed be $v_s$ with respect to the surface.

Since they are initially at rest on a frictionless surface, there is no external horizontal force on the system.

Hence, horizontal momentum is conserved: $50v_m + 20v_s = 0$

Choose directions

Let the man move in the positive direction after pushing. Then the son moves in the opposite direction.

So, $50v_m = 20v_s$ in magnitude, i.e. $v_s = \frac{50}{20}v_m = 2.5v_m$

Use the given relative speed

The son moves at speed $0.70\,\text{m s}^{-1}$ with respect to the man.

Since they move in opposite directions, relative speed is the sum of their speeds: $v_s + v_m = 0.70$

Substitute $v_s = 2.5v_m$: $2.5v_m + v_m = 0.70$ $3.5v_m = 0.70$ $v_m = \frac{0.70}{3.5} = 0.20\,\text{m s}^{-1}$

Match with options

$v_m = 0.20\,\text{m s}^{-1}$

So the correct option is C.

Verification with stored answer

Stored correct answer: C

Our derived answer: C

Hence, they agree.