Convince yourself that the higher the units digit and tens digit, the smaller the ratio will be. Then convince yourself that a higher hundreds digit will make the ratio bigger. From here, it is clear that the number must be $199$, and the solution is $|1-9| = 8$ hence (b).
Why you want the units digit to be higher:
Let the number be $x$ and the digital sum be $d$.
Increase the units digit by $1$ (without changing the hundreds digit).
Then $x/d > (x+10)/(d+1)$ iff $xd + x > xd + 10d$ iff $x > 10d$.
If $x = 100a + 10b + c$ is a three digit number, then:
$x > 10d$ iff $100a + 10b + c > 10a + 10b + 10c$ iff $90a - 9c > 0$ iff $9(10a - c) > 0$.
Since $a\neq0$ and $c$ are digits, we know that $10a > c$; hence the last inequality holds, so that the new ratio is, indeed, smaller. A similar analysis can be carried out for the other digit places in the event that "convince yourself" is not convincing enough.