Solution:

The time taken by an object to fall from a height is given by the equation:

\[
t = \sqrt{\frac{2h}{g}}
\]

For mass \( m_a \) dropped from height \( a \), the time taken is:

\[
t_a = \sqrt{\frac{2a}{g}}
\]

For mass \( m_b \) dropped from height \( b \), the time taken is:

\[
t_b = \sqrt{\frac{2b}{g}}
\]

The ratio of time taken by the two bodies is:

\[
\frac{t_a}{t_b} = \frac{\sqrt{\frac{2a}{g}}}{\sqrt{\frac{2b}{g}}} = \sqrt{\frac{a}{b}}
\]

So, the ratio of the time taken is:

\[
\frac{t_a}{t_b} = \sqrt{\frac{a}{b}}
\]