The magnetic force, charge, and velocity of a particle are given. I am asked to find the magnetic field.
$$q = -2 \,\mathrm{\mu C}$$ $$\vec{v} = (-\hat i + 3\hat j) \times 10^6 \mathrm{m/s}$$ $$\vec{F} = (3\hat i + \hat j + 2 \hat k) \mathrm{N}$$ $$B_{x} = 0$$
Use the equation of Lorentz force to calculate the field vector.
Quoting from this link,
If a particle of charge $q$ moves with velocity $v$ in the presence of an electric field $E$ and a magnetic field $B$, then it will experience a force $$\mathbf{F} = q\left[\mathbf{E} + (\mathbf{v} \times \mathbf{B})\right]$$
In your case, there is no electric field. So calculate accordingly.
Given: $B_x=0$
Take $B_y$ and $B_z$ to be the other components. Calculate the cross product and then compare the coefficients of the unit vectors, to get the answer.
Hope it helps.
Let $\vec{B}=B_y\hat {j}+B_z\hat k$ (since, $B_x=0$) be the magnetic field then the magnetic force ($F_m$) acting on the charge, is given as $$F_m=q(\vec v\times \vec B)$$ setting the corresponding values of $\vec F$, $q$, $\vec v$ & $\vec B$, $$3\hat i+\hat j+2\hat k=2\times 10^{-6}\times 10^{6}((-\hat i+3\hat j))(B_y\hat {j}+B_z\hat k)$$ $$3\hat i+\hat j+2\hat k=2(3B_z\hat i+B_z\hat k-B_y\hat k)$$ $$3\hat i+\hat j+2\hat k=6B_z\hat i+2B_z\hat k-2B_y\hat k$$ Now, compare the corresponding coefficients on both the sides, to find $B_y$ & $B_z$ & then the magnetic field $\vec B$
