1. Given electric fields

$\vec E_1=E_0\hat j\cos(\omega t-kx),\qquad \vec E_2=E_0\hat k\cos(\omega t-ky)$

At the origin and at $t=0$:

$\cos(0)=1$

So,

$\vec E_1=E_0\hat j,\qquad \vec E_2=E_0\hat k$

Hence total electric field is

$\vec E=E_0(\hat j+\hat k)$

2. Find magnetic fields of the waves

For an electromagnetic wave in vacuum,

$\vec B=\frac{1}{c}(\hat n\times \vec E)$

where $\hat n$ is the propagation direction.

For wave 1:

The phase is $\omega t-kx$, so it propagates along $+x$ direction. Thus $\hat n=\hat i$.

$\vec B_1=\frac{1}{c}(\hat i\times E_0\hat j)=\frac{E_0}{c}\hat k$

For wave 2:

The phase is $\omega t-ky$, so it propagates along $+y$ direction. Thus $\hat n=\hat j$.

$\vec B_2=\frac{1}{c}(\hat j\times E_0\hat k)=\frac{E_0}{c}\hat i$

At origin and $t=0$,

$\vec B=\vec B_1+\vec B_2=\frac{E_0}{c}(\hat i+\hat k)$

3. Given particle velocity

$\vec v=0.8c\hat j$

Lorentz force is

$\vec F=q(\vec E+\vec v\times \vec B)$

So first compute $\vec v\times \vec B$:

$\vec v\times \vec B=(0.8c\hat j)\times \frac{E_0}{c}(\hat i+\hat k)$

$=0.8E_0\left(\hat j\times \hat i+\hat j\times \hat k\right)$

Using cross products,

$\hat j\times \hat i=-\hat k,\qquad \hat j\times \hat k=\hat i$

Therefore,

$\vec v\times \vec B=0.8E_0(\hat i-\hat k)$

4. Total force

$\vec F=q\left[E_0(\hat j+\hat k)+0.8E_0(\hat i-\hat k)\right]$

$\vec F=E_0q\left(0.8\hat i+\hat j+(1-0.8)\hat k\right)$

$\vec F=E_0q\left(0.8\hat i+\hat j+0.2\hat k\right)$

5. Compare with options

This matches:

$\boxed{\text{C: }E_0q\left(0.8\hat i+\hat j+0.2\hat k\right)}$

So the derived answer agrees with the stored correct answer.