Answer
$1.2\;\rm m/s$
Work Step by Step
As you see in the figures below, the net force exerted on the bar must be zero in both directions $x'$ and $y'$, so the bar will move at a constant velocity down the ramp.
The terminal speed occurs when the net force exerted on the bar in the $x'$-direction is zero.
$$\sum F_{x'}=F_G\sin6^\circ-F_B\cos6^\circ=ma_x=m(0)=0$$
$$F_G\sin6^\circ=F_B\cos6^\circ $$
$$mg\sin6^\circ=IlB\cos6^\circ \tag 1$$
Noting that there will be an induced $ \rm emf$ due to the motion of the bar in the magnetic field.
This induced voltage is given by
$$\varepsilon=Blv\cos6^\circ$$
whereas $B\cos6^\circ$ is the magnetic field component which is perpendicular to the bar's velocity.
And hence, the induces current is given by Ohm's law
$$\varepsilon=IR$$
$$I=\dfrac{\varepsilon}{R}=\dfrac{Blv\cos6^\circ}{R}$$
Plugging into (1);
$$mg\sin6^\circ=\dfrac{Blv\cos6^\circ}{R}\;lB\cos6^\circ $$
$$mg\sin6^\circ=\dfrac{B^2l^2v\cos^26^\circ}{R} $$
Solving for $v$;
$$v=\dfrac{mgR\sin6^\circ}{B^2l^2 \cos^26^\circ} $$
Plugging the known;
$$v=\dfrac{0.04\cdot 9.8\cdot 0.6
\cdot \sin6^\circ}{0.45^2\cdot 0.32^2 \cdot\cos^26^\circ} $$
$$v=\color{red}{\bf 1.2}\;\rm m/s$$
