1. Intensity after first polarizer $A$

   Unpolarized light of intensity $I$ passes through an ideal polarizer. The transmitted intensity becomes $I_A = \frac{I}{2}.$

2. Using the given intensity after polarizer $B$

   Let the angle between the transmission axes of $A$ and $B$ be $\theta$.

   By Malus' law, $I_B = I_A \cos^2\theta = \frac{I}{2}\cos^2\theta.$

   Given that intensity beyond $B$ is $\frac{I}{2}$, $\frac{I}{2}\cos^2\theta = \frac{I}{2}.$

   Hence, $\cos^2\theta = 1 \implies \theta = 0^\circ.$

   So, polarizers $A$ and $B$ are parallel.

3. Insert polarizer $C$ between $A$ and $B$

   Let the angle between $A$ and $C$ be $\phi$.

   Since $A$ and $B$ are parallel, the angle between $C$ and $B$ is also $\phi$.

   After passing through $C$: $I_C = I_A \cos^2\phi = \frac{I}{2}\cos^2\phi.$

   After then passing through $B$: $I_{final} = I_C \cos^2\phi = \frac{I}{2}\cos^4\phi.$

   Given: $\frac{I}{2}\cos^4\phi = \frac{I}{8}.$

   Therefore, $\cos^4\phi = \frac{1}{4}.$

   Taking square root, $\cos^2\phi = \frac{1}{2}.$

   Hence, $\phi = 45^\circ.$

4. Check options

   • A: $60^\circ$ $\rightarrow$ incorrect
   • B: $30^\circ$ $\rightarrow$ incorrect
   • C: $45^\circ$ $\rightarrow$ correct
   • D: $0^\circ$ $\rightarrow$ incorrect

Therefore, the angle between polarizer $A$ and $C$ is $\boxed{45^\circ}$.