1. Forces along the incline

The rod slides down the smooth rails, which are inclined at angle $\theta$ to the horizontal.

So the component of gravity along the rails is $F_g = mg\sin\theta.$

2. Motional emf induced in the rod

If the rod moves with speed $v$ along the rails in a magnetic field $B$ normal to the plane of the rails, then the motional emf across the rod is $\mathcal E = Blv.$

3. Current in the circuit

The rails and rod form a closed circuit with resistance $R$, so the induced current is $I = \frac{\mathcal E}{R} = \frac{Blv}{R}.$

4. Magnetic force opposing motion

The magnetic force on a current-carrying rod of length $l$ in field $B$ is $F_m = BIl.$

Substituting $I$, $F_m = B\left(\frac{Blv}{R}\right)l = \frac{B^2l^2v}{R}.$

This force acts up the incline, opposing the downward motion.

5. Condition for terminal speed

At terminal speed, acceleration becomes zero, so net force along the incline is zero: $mg\sin\theta = \frac{B^2l^2v_t}{R}.$

Solving for $v_t$, $v_t = \frac{mgR\sin\theta}{B^2l^2}.$

6. Matching with options

This corresponds to $\boxed{\frac{mgR\sin\theta}{B^2l^2}}$ which is Option C.