Answer:
Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.
Step-by-step explanation:
Considering the complete question attached in figure below.
Time period for balance wheel is:


m = mass of balance wheel
R = radius of balance wheel.
Angular frequency is related to Time period as:

As dimensions of new balance wheel are one-third of their original values

