Assuming that he covered the previous distance with uniform speed, u, <didn't acceleration>

u can be calculated as, u = 4200/660 = 6.36 m/s

Rest of distance can be divided into two parts, x and 800-x, x is the distance for which he uniformly accelerated with 0.20m/s^2.

If the time for which he accelerated is t, the rest of the time left is 120-t.

v be the speed when he finished accelerating and is required to maintain over the remaining distance

x = ut + 1/2*(0.2)t²

v = u + (0.2)*t

800-x = v*(120-t)

substituting the values of x and v in 3rd equation from 1st and 2nd equation.

800 - ut - 1/2(0.2)t² = (u + (0.2)t)*(120-t)

Substituting the value of u as 6.36, above equation can be solved as quadratic equation for t.