Of the 3 numbers whose average is $64$, the first number is $\frac{1}{3}$ times the sum of the other 2. The first number is:

Let the three numbers be $a$, $b$, and $c$.The average of the three numbers is $64$:
$$\frac{a+b+c}{3}=64⟹a+b+c=192$$
We are given that the first number ($a$) is $\frac{1}{3}$ times the sum of the other two ($b+c$):
$$a=\frac{1}{3}(b+c)⟹b+c=3a$$
Substitute $b+c=3a$ into the total sum equation:
$$a+3a=192$$
$$4a=192$$
$$a=\frac{192}{4}=48$$