Given, speed of proton, $v=5 \times 10^6 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$ Electric field, $\mathbf{E}=4 \times 10^6[2 \hat{\mathbf{i}}+0.2 \hat{\mathbf{j}}+0.1 \hat{\mathbf{k}}] \mathrm{V} / \mathrm{m}$ Magnetic field, $\mathbf{B}=0.2[\hat{\mathbf{i}}+0.2 \hat{\mathbf{j}}+\hat{\mathbf{k}}] \mathrm{T}$ Charge on proton, $q=1.6 \times 10^{-19} \mathrm{C}$
Net force acting on the proton is calculated according to Lorentz's force as

$$
\begin{aligned}
& \mathbf{F}=q[\mathbf{E}+\mathbf{v} \times \mathbf{B}] \\
& =1.6 \times 10^{-19}\left[4 \times 10^6(2 \hat{\mathbf{i}}+0.2 \hat{\mathbf{j}}+0.1 \hat{\mathbf{k}})\right. \\
& \left.+5 \times 10^6 \hat{\mathbf{j}} \times 0.2(\hat{\mathbf{i}}+0.2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\right] \\
& =1.6 \times 10^{-19} \times 10^6[4(2 \hat{\mathbf{i}}+0.2 \hat{\mathbf{j}}+0.1 \hat{\mathbf{k}}) \\
& +(-\hat{\mathbf{k}}+0+5 \hat{\mathbf{i}})] \\
& \Rightarrow \mathbf{F}=1.6 \times 10^{-13}[13 \hat{\mathbf{i}}+0.8 \hat{\mathbf{j}}-0.6 \hat{\mathbf{k}}] \\
& \therefore \mathbf{F}=|\mathrm{F}|=1.6 \times 10^{-13} \sqrt{(13)^2+(0.8)^2+(0.6)^2} \\
& =1.6 \times 10^{-13} \sqrt{170}=20.86 \times 10^{-13} \mathrm{~N} \\
& =20 \times 10^{-13} \mathrm{~N} \\
&
\end{aligned}
$$