Answer
a) ${\bf 1}\;\rm N$
b) ${\bf 2.24}\;\rm T$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the pulling power is given by
$$P_{\rm input}=F_{\rm pull}v=\dfrac{v^2l^2B^2}{R}\tag 1$$
So, the pulling force is
$$F_{\rm pull}=\dfrac{P_{\rm input}}{v}$$
Plug the known;
$$F_{\rm pull}=\dfrac{ 4}{4}=\color{red}{\bf 1}\;\rm N$$
$$\color{blue}{\bf [b]}$$
The strength of the magnetic field is given by solving (1) for $B$,
$$B=\sqrt{\dfrac{F_{\rm pull} R}{vl^2}}$$
Plug the known;
$$B=\sqrt{\dfrac{(1)(0.2)}{(4)(0.10)^2}}$$
$$B=\color{red}{\bf 2.24}\;\rm T$$
