Answer
$2.7\times 10^{-15}N$
Work Step by Step
We can find the required net magnetic force as follows:
$\vec F_e=e\vec E$
$\vec F_e=(1.6\times 10^{-19}C)(220N/C)=(3.52\times 10^{-17}N)\hat x$
and $\vec F_B=e(\vec V\times \vec B)$
$\implies \vec F_B=1.6\times 10^{-19}[(0.67\times 10^{-4})(-0.11T)\hat x-(1.5\times 10^{5})(-0.11)\hat y-(0.67\times 10^4)(0.25T)\hat z]$
$\implies \vec F_B=(-1.18\times 10^{-16}N)\hat x=(2.64\times 10^{-15}N)\hat y-(2.68\times 10^{-16}N)\hat z$
Now $\vec{F_{net}}=\vec{F_E}+\vec{F_B}$
We plug in the known values to obtain:
$\vec{F_{net}}=(-8.3\times 10^{-17})\hat x+(2.64\times 10^{-15}N)\hat y-(2.68\times 10^{-16}N)\hat z$
$\implies |F_{net}|=\sqrt{F_x^2+F_y^2+F_Z^2}$
$|F_{net}|=\sqrt{(-8.3\times 10^{-7})^2+(2.64\times 10^{-15})^2+(-2.68\times 10^{-16})^2}$
$|F_{net}|=2.7\times 10^{-15}N$
