$K$ x-rays are released when an electron transitions to an $n = 1$ orbital from a higher orbital. The energy released is
$\begin{align*}\frac{\mu Z^2 e^4}{2 \hbar^2 \left(4 \pi \epsilon_0 \right)^2}\left(1 - \frac{1}{m^2} \right)\end{align*}$
where $m$ is the shell of the higher orbital. We are given that $Z = 6$ for carbon and $Z = 12$ for magnesium, so the desired ratio is
$\begin{align*}\left(\frac{6}{12}\right)^2 = \boxed{\frac{1}{4}}\end{align*}$
Therefore, answer (A) is correct.

Solution to 1992 Problem 3