Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are the initial and final velocities, respecitvely; [tex]a[/tex] is the acceleration; and [tex]\Delta x[/tex] is the change in position.
So we have
[tex]\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)[/tex]
[tex]\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}[/tex]
(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)
