Answer
$B = 1.5\cdot10^{-3}$ T
$\theta = 30^{\circ}$
Work Step by Step
Recall that the magnetic force on a moving particle is $\vec{F}_B = q\vec{v} \times \vec{B} = q\vec{v}\vec{B}$
Because both protons are going in two different directions, we can calculate the magnitude of $\vec{B}$ such that
$|\vec{B}| = \sqrt{\displaystyle (\frac{\vec{F}_{B,1}}{q\vec{v}_1})^2+ (\frac{\vec{F}_{B,2}}{q\vec{v}_2})^2}$
$B = \sqrt{\displaystyle (\frac{\vec{F}_1 \hat k}{q\vec{v}_1\hat i})^2 + (\frac{\vec{F}_2\hat k}{q\vec{v}_2\hat j})^2}$
When the unit vectors are squared, they will cancel out (the dot product of the same unit vector is 1)
$B = \sqrt{\displaystyle (\frac{1.20 \cdot 10^{-16} }{(1.6\cdot10^{-19})(10^6)})^2 + (\frac{-4.16\cdot 10^{-16}}{(1.6\cdot10^{-19})(2.00\cdot10^6)})^2}$
$B = 1.5\cdot10^{-3}$ T
The angle clockwise to the +$x$-axis is given as
$\displaystyle \tan(\theta_{cw}) = \frac{B_y}{B_x}$
$\displaystyle \theta_{cw} = \tan^{-1}(\frac{B_y}{B_x}) = \tan^{-1}(\frac{\frac{\vec{F}_1}{q\vec{v}_1}}{\frac{\vec{F}_2}{q\vec{v}_2}}) = \tan^{-1}(\frac{\frac{\vec{F}_1}{\vec{v}_1}}{\frac{\vec{F}_2}{\vec{v}_2}}) = \tan^{-1}(\frac{\vec{F}_1\vec{v}_2}{\vec{v}_1\vec{F}_2})$
$\displaystyle \theta_{cw} = \tan^{-1}(\frac{(1.20\cdot10^{-16})(2.00\cdot10^6)}{(10^6)(-4.16\cdot10^{-16})}) \approx -30^{\circ}$
This makes the angle counterclockwise to the +$x$-axis
$\theta_{ccw} \approx 30^{\circ}$
