The correct answer is $\boxed{\frac{1}{3}}$.
A probability density function (PDF) is a function that gives the probability of a continuous random variable taking on a value within a given interval. The total area under the curve of a PDF must be equal to 1.
In the given figure, the graph of the function $f(x)$ is a triangle with base $3$ and height $1$. The area of a triangle is equal to $\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. Therefore, the area of the triangle in the figure is $\frac{1}{2}(3)(1) = \frac{3}{2}$.
In order for $f(x)$ to be a valid PDF, the total area under the curve must be equal to 1. Since the area of the triangle is $\frac{3}{2}$, the value of $h$ must be such that $\frac{1}{2}(3)(h) = 1$. Solving for $h$, we get $h = \frac{1}{3}$.
Therefore, the correct answer is $\boxed{\frac{1}{3}}$.
The other options are incorrect because they do not satisfy the condition that the total area under the curve of a PDF must be equal to 1.