Answer
$B_x=-2.0T$
Work Step by Step
We know that
$F=q(v\times B)$
We plug in the known values to obtain:
$(6.4\times10^{-19})k^{\wedge}=(-1.6\times10^{-19})(2i^{\wedge}+4j^{\wedge})\times(B_xi^{\wedge}+3B_xj^{\wedge})$
We know that $i^{\wedge}\times i^{\wedge}=j^{\wedge}\times j^{\wedge}=k^{\wedge}\times k^{\wedge}=0$ and $i^{\wedge}\times j^{\wedge}=K^{\wedge}$, $j^{\wedge}\times i^{\wedge}=-K^{\wedge}$
Therefore,
$(6.4\times10^{-19})k^{\wedge}=(-1.6\times10^{-19})(6B_xk^{\wedge}+4B_x(-k^{\wedge}))$
$(6.4\times10^{-19})k^{\wedge}=(-1.6\times10^{-19})(2B_xk^{\wedge})$
$(6.4\times10^{-19})=(-1.6\times10^{-19})(2B_x)$
$B_x=-2.0T$
