1. Given data

   • Refractive index of glass: $n=1.5$
   • Reflected light intensity fraction: $R=4\%=0.04$
   • Incident electric field amplitude: $E_i=30\,\text{V/m}$

2. Relate intensity and electric field amplitude

   For an electromagnetic wave, intensity is proportional to the square of the electric field amplitude: $I \propto E^2$

   Hence, if $4\%$ of the light is reflected, then the reflected field amplitude satisfies $\frac{I_r}{I_i}=\left(\frac{E_r}{E_i}\right)^2=0.04$

   Therefore, $\frac{E_r}{E_i}=\sqrt{0.04}=0.2$ $E_r=0.2\times 30=6\,\text{V/m}$

3. Use boundary condition for normal incidence

   At the boundary, the tangential component of electric field is continuous, so $E_i+E_r=E_t$ where $E_t$ is the transmitted electric field amplitude in glass.

   Thus, $E_t=30+(-6)$

   Since reflection from a denser medium introduces a phase change of $\pi$, the reflected electric field is opposite in sign. So effectively, $E_t=30-6=24\,\text{V/m}$

4. Final answer $E_t=24\,\text{V/m}$

5. Option check

   • A: $6\,\text{V/m}$ $\rightarrow$ reflected amplitude, not transmitted
   • B: $10\,\text{V/m}$ $\rightarrow$ incorrect
   • C: $30\,\text{V/m}$ $\rightarrow$ incorrect
   • D: $24\,\text{V/m}$ $\rightarrow$ correct

Therefore, the amplitude of the electric field in the glass medium is $24\,\text{V/m}$.