How is the property in eq 15 obtained for Xavier initialization
The fixed answer of $1/3$ is a result of their decision to use the uniform distribution along with the parameterised arguments, namely $1/\sqrt{n}$.

For a uniform distribution, denoted with lower and upper bounds $a$ and $b$:

$$
U(a, b)
$$

the variance is defined as:

$$
\frac{1}{12} (b - a)^2
$$

So in the case of the authors, Glorot and Bengio, the two bounds are simply the square root of the number of neurons in the layer of interest (generally referring to the preceding layer, as they put it). This size is called $n$, and they set the bounds on the uniform distribution as:

$$
a = - \frac{1}{\sqrt{n}}
$$
$$
b = \frac{1}{\sqrt{n}}
$$

So if we plug these values into equation 15, we get:

$$
Var = \frac{1}{12}(\frac{1}{\sqrt{n}} - -\frac{1}{\sqrt{n}})^2
$$

$$
Var = \frac{1}{12}(\frac{2}{\sqrt{n}})^2
$$

$$
Var = \frac{1}{12} * \frac{4}{n}
$$

And so finally:

$$
n * Var = \frac{1}{3}
$$