The sum of three numbers is 132. If the first number be twice the second and third number is one third of the first, then the second number is

Let's denote the three numbers as $A$, $B$, and $C$.

1. According to the problem, the sum of the three numbers is 132: $A + B + C = 132$

2. The first number $A$ is twice the second number $B$: $A = 2B$

3. The third number $C$ is one third of the first number $A$: $C = \frac{A}{3}$

Now, we substitute the expressions for $A$ and $C$ into the sum equation:

4. Substitute $A = 2B$ and $C = \frac{2B}{3}$ into $A + B + C = 132$: $2B + B + \frac{2B}{3} = 132$

5. Combine the terms on the left side: $2B + B + \frac{2B}{3} = 132$ $\left(2 + 1 + \frac{2}{3}\right)B = 132$ $\left(\frac{3 \cdot 2 + 3 \cdot 1 + 2}{3}\right)B = 132$ $\left(\frac{6 + 3 + 2}{3}\right)B = 132$ $\left(\frac{11}{3}\right)B = 132$

6. Solve for $B$: $B = 132 \times \frac{3}{11}$ $B = 12 \times 3$ $B = 36$

Therefore, the second number is $36$.