D

Step 1: Understand the relationship between voltage and intensity in an X-ray tube. The intensity of X-rays produced in an X-ray tube is given by the formula:
$$ I \propto V^n $$
where $I$ is the intensity, $V$ is the voltage, and $n$ is a constant that varies depending on the specific setup, typically $n = 2$.
Step 2: If the voltage is doubled ($V' = 2V$), then the new intensity can be calculated as follows:
$$ I' \propto (2V)^n = 2^n \cdot V^n $$
Step 3: Substituting $n = 2$ gives us:
$$ I' \propto 2^2 \cdot V^n = 4 \cdot V^n $$
Step 4: The new intensity $I'$ is therefore 4 times the original intensity $I$. Thus, if the voltage is doubled, the intensity will become four times greater.
Therefore, the correct answer is: D.