Answer
${\bf 1.50}\;\rm mT\tag {$30^\circ$ CCW from $+x$}$
Work Step by Step
According to the right-hand rule, the magnetic field exerted on the first proton is in the positive $y$-direction. And by the same rule, the direction of the magnetic field exerted on the second proton is in the positive $x$-direction.
This means that the direction of the magnetic is between the positive $x$-direction and the positive $y$-direction. In other words, the direction of the magnetic field is in the first quadrant of the $x$-$y$ plane.
Recalling that the magnetic force exerted on a moving charge is given by
$$F=qvB\sin\theta$$
Hence, $F_1=ev_1B\sin\theta_1$, and $F_2=ev_2B\sin\theta_2$
and since the
where $\theta_1$ and $\theta_2$ is the angle between the magnetic field and the speeds of the protons. The first proton is to the right in the $x$-$y$ plane, so the angle between $B$ and $v_1$ is $\theta$, while the angle between $B$ and $v_2$, since it is upward in the same place, is $90^\circ-\theta$, see the figure below.
Now we need to find $\theta$, so
$$\dfrac{F_1}{F_2}=\dfrac{ev_1B\sin\theta}{ev_2B\sin(90^\circ-\theta)}= \dfrac{ v_1 \sin\theta}{ v_2 \cos\theta}=\dfrac{ v_1 \tan\theta}{ v_2 }$$
Hence,
$$\theta=\tan^{-1}\left[ \dfrac{F_1v_2}{F_2v_1}\right]$$
Plug the given;
$$\theta=\tan^{-1}\left[ \dfrac{(1.20 \times 10^{-16})(2\times 10^6)}{(4.16 \times 10^{-16})(1\times 10^6)}\right]=\bf 29.89^\circ\approx 30^\circ$$
Thus, the magnitude of the magnetic field is given by
$$B=\dfrac{F_1}{ev_1\sin30^\circ}=\dfrac{(1.20 \times 10^{-16})}{(1.6\times 10^{-19})(1\times 10^6)\sin30^\circ}$$
$$B=\color{red}{\bf 1.50}\;\rm mT$$
