Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The wire is moving at a constant speed which means that the pushing force equals the magnetic force exerted on this wire.
$$F_{\rm push}=F_B$$
We know that the magnetic force exerted on a moving wire is given by
$$F_{\rm push} =BIL\tag 1$$
where $I$ is the induced current and we need to find it.
To find the induced current, we must find the induced emf first.
The induced emf for a wire that moves in a magnetic field is given by
$$\varepsilon=vLB$$
The induced current is given by
$$I=\dfrac{\varepsilon}{R}$$
$$I=\dfrac{vLB}{R}\tag 2$$
Plug into (1),
$$F_{\rm push} =BL\dfrac{vLB}{R}$$
$$F_{\rm push} =\dfrac{B^2L^2 v}{R}$$
Plug the known;
$$F_{\rm push} =\dfrac{(0.50)^2(0.1)^2 (0.50)}{(2)}$$
$$F_{\rm push} =\color{red}{\bf 6.25\times 10^{-4}}\;\rm N$$
$$\color{blue}{\bf [b]}$$
Recalling that the power is given by
$$P=Fv=F_{\rm push} v$$
Plug the known;
$$P =(6.25\times 10^{-4})(0.50)$$
$$P =\color{red}{\bf 3.125\times 10^{-4}}\;\rm W$$
$$\color{blue}{\bf [c]}$$
The exposed area, to the magnetic flux, of the loop decreases while the wire is moving to the right (the area of the closed loop itself decreases) which means that the flux is decreasing. According to Lenz's law, the induced current must generate a flux that fights this decrease. And since the flux is out of the page the induced flux must also be out of the page. Hence the direction of the current, according to the right-hand rule, must be counterclockwise.
The magnitude of the induced current is given by (2),
$$I=\dfrac{vLB}{R}=\dfrac{(0.50)(0.10)(0.50)}{(2)}$$
$$I=\color{red}{\bf 12.5}\;\rm mA$$
$$\color{blue}{\bf [d]}$$
We know that the power dissipated in the resistor is given by
$$P=I^2R$$
Plug the known;
$$P=(12.5\times 10^{-3})^2(2)$$
$$P=\color{red}{\bf 3.125\times 10^{-4}}\;\rm W$$
