A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198, then the difference between the first and last digits is

The correct option is 2.

Let us represent the three-digit number. Let the hundreds digit be represented by $x$ and the units digit by $y$. The question states that the middle digit (the tens digit) is 0. Thus, the three-digit number can be written in the form $x0y$.

In place-value notation, the value of this number is:
$100x+0·10+y=100x+y$

We are given that the sum of the digits is 4. Therefore:
$x+0+y=4$
Which simplifies to:
$x+y=4$ (Equation 1)

Next, the number formed by interchanging the first (hundreds) and last (units) digits is $y0x$. The value of this new number is:
$100y+x$

According to the problem, this interchanged number is greater than the original number by 198. We can write this relation as:
$(100y+x)-(100x+y)=198$

Simplifying the left-hand side of the equation:
$100y+x-100x-y=198$
$99y-99x=198$

Dividing both sides of the equation by 99:
$y-x=2$ (Equation 2)

The problem asks for the difference between the first and last digits, which is the absolute difference between $x$ and $y$, represented as $|y-x|$. From Equation 2, we directly find that:
$y-x=2$

Thus, the difference between the first and last digits is 2.