The initial height on each side was $20 \mbox{ cm}$ and $5 \mbox{ cm}$ was added to one side, so the sum of the heights on the two sides must be
$\begin{align}h_1 + h_2 = 45 \mbox{ cm} \label{eqn:1}\end{align}$
Also, the pressure $5 \mbox{ cm}$ below the top of the left column must equal to the pressure at that same height on the right column. Therefore, the weight of the fluid above each of these locations must be the same
$\begin{align*}5 \mbox{ cm} \cdot 4 \;\mathrm{g}\mathrm{/}\mathrm{cm}^3 \cdot g \cdot A = (h_2 - h_1 + 5 \mbox{ cm}) \cdot 1 \...$
where $g$ is the acceleration of gravity and $A$ is the cross-sectional area of the tube. The above equation reduces to
$\setcounter{equation}{1}\begin{align}h_2 - h_1 = 15 \mbox{ cm} \label{eqn:2}\end{align}$
The solution to equations (1) and (2) is
$\begin{align*}h_1 &= 15 \mbox{ cm} \\h_2 &= 30 \mbox{ cm}\end{align*}$
Therefore, the desired ratio is
$\begin{align*}\frac{h_2}{h_1} = \frac{30 \mbox{ cm}}{15 \mbox{ cm}} = \boxed{2}\end{align*}$
Therefore, answer (C) is correct.

Solution to 1996 Problem 30