As we know that,
T = 2 \pi \sqrt{l\over g}
Time period is directly proportional to the square root of the length of the pendulum.
In the case when the length is 1 m,
T1 = 2 \pi \sqrt{1\over g}
and
In the case when the length is 9m,
T2 = 2 \pi \sqrt{9\over g}
So, a comparison of T1 and T2 gives
T1 : T2 = 2 \pi \sqrt{l\over g} : 2 \pi \sqrt{9\over g}
⇒ T1 : T2 = \sqrt{1\over 9}
⇒ T1 : T2 = 1\over 3
Hence, T1 : T2 = 1 : 3