A straight wire of mass m and length l is able to slide on two...

Since the rails are made of plastic, they are non-conducting. Therefore, the current $i$ must flow through the entire length $l$ of the wire, not just the part between the rails.

Let the magnetic field $B$ be applied vertically to the horizontal plane. The magnetic force acting on the current-carrying wire is given by:
$F_m = i l B$

The normal reaction force on the wire due to its weight is:
$N = m g$

The maximum static frictional force opposing the motion of the wire is:
$f = \mu N = \mu m g$

For the wire to just start sliding on the rails, the magnetic force must be equal to or greater than the maximum static frictional force:
$F_m \ge f$

Substituting the expressions for $F_m$ and $f$:
$i l B \ge \mu m g$

Rearranging the inequality to solve for the minimum magnetic field $B$:
$B \ge \dfrac{\mu m g}{i l}$

Thus, the minimum magnetic field required to make the wire slide is $\dfrac{\mu m g}{i l}$.