Answer
a) $({\bf 5.66\times 10^{-13}}{\;\rm N})\;\hat j$
b) $({\bf 8\times 10^{-13}}{\;\rm N})\;\hat k$
Work Step by Step
We know that the force exerted by a magnetic field on a moving charge is given by
$$F=q\vec v\times \vec B\tag 1$$
$$\color{blue}{\bf [a]}$$
Plug the known from the given graph into (1)
$$F=(1.6\times 10^{-19})(1\times 10^7)(\cos45^\circ\;\hat i+0\;\hat j+\sin45^\circ\;\hat k)(0.5\;\hat i)$$
As we see, the velocity has a zero component in the $y$-direction.
Thus,
$$F= (\color{red}{\bf 5.66\times 10^{-13}}{\;\rm N})\;\hat j$$
$$\color{blue}{\bf [b]}$$
Plug the known from the given graph into (1)
$$F=(1.6\times 10^{-19})(1\times 10^7)(0\;\hat i-1\;\hat j+0\;\hat k)(0.5\;\hat i)$$
As we see, the velocity has one component in the negative $y$-direction.
Thus,
$$F= (\color{red}{\bf 8\times 10^{-13}}{\;\rm N})\;\hat k$$
