1. Use Malus' law

For plane polarized light, the intensity after a polarizer is

$I = I_0 \cos^2(\phi - \theta)$

where:

• $\phi$ = angle of polarization of incident light with the $x$-axis,
• $\theta$ = angle of the pass axis of the polarizer with the $x$-axis.

We are told that the observed intensity is the same for

$\theta = 8^\circ,\ 38^\circ,\ 188^\circ,\ 218^\circ.$

So,

$\cos^2(\phi-8^\circ)=\cos^2(\phi-38^\circ)=\cos^2(\phi-188^\circ)=\cos^2(\phi-218^\circ).$

2. Use periodicity of $\cos^2$

Since

$\cos^2(x+180^\circ)=\cos^2 x,$

the equalities for $188^\circ$ and $218^\circ$ are automatically the same as those for $8^\circ$ and $38^\circ$ respectively. So the essential condition is

$\cos^2(\phi-8^\circ)=\cos^2(\phi-38^\circ).$

3. Solve the condition $\cos^2 A = \cos^2 B$

This happens when either

$A = B + n\cdot 180^\circ$

or

$A = -B + n\cdot 180^\circ.$

Here,

$A=\phi-8^\circ, \qquad B=\phi-38^\circ.$

The first case gives

$(\phi-8^\circ)=(\phi-38^\circ)+n\cdot180^\circ$ $30^\circ=n\cdot180^\circ,$ which is impossible.

So use the second case:

$(\phi-8^\circ)=-(\phi-38^\circ)+n\cdot180^\circ$

$\phi-8^\circ=-\phi+38^\circ+n\cdot180^\circ$

$2\phi=46^\circ+n\cdot180^\circ$

$\phi=23^\circ+n\cdot90^\circ.$

Thus possible polarization directions are

$\phi=23^\circ,\ 113^\circ,\ 203^\circ,\ 293^\circ.$

4. Match with options

Among the given options:

• $98^\circ$ — not possible
• $128^\circ$ — not possible
• $203^\circ$ — possible
• $45^\circ$ — not possible

Therefore, the required angle is

$\boxed{203^\circ}$

So the correct option is C.