A conducting rod AB is in contact with metal rails AC and BD, which...

The motional emf induced in the moving rod is given by:
$E = Bvl$

Substituting the given values ($B = 0.4\text{ T}$, $v = 5\text{ m/s}$, $l = 0.25\text{ m}$):
$E = 0.4 \times 5 \times 0.25 = 0.5\text{ V}$

The magnitude of the induced current is:
$I = \dfrac{E}{R}$
$I = \dfrac{0.5}{5} = 0.1\text{ A}$

The direction of the induced current can be determined using Fleming's Right-Hand Rule or the Lorentz force on positive charges ($\vec{F} = q(\vec{v} \times \vec{B})$).
With velocity $\vec{v}$ towards the right and magnetic field $\vec{B}$ into the plane of the paper, the force on positive charges is directed upwards.
Therefore, the induced current in the rod AB flows from B to A.

Answer: $0.1$ A from B to A