Let the three-digit number be XYZ.

This number is such that $\frac{100X + 10Y + Z}{X + Y + Z}$ is the least possible.

For 100, this ratio is $100/1 = 100$ For 101, this ratio is $101/2 = 50.5$ For 109, this ratio is $109/10 = 10.9$ For 110, this ratio is $110/2 = 55$ For 119, this ratio is $119/11 = 10.81$ For 129, this ratio is $129/12 = 10.75$

So, we can see that in 100-199 range, this ratio will be the least for 199, which is $199/19 = 10.47$

Similarly, in 200-299 range, this ratio will be the least for 299, which is $299/20 = 14.95$

Similarly, in 300-399 range, this ratio will be the least for 399, which is $399/21 = 19$

We can see this ratio is slowly increasing. For 999, it would be $999/27 = 37$

So, this ratio is the least for 199.

The difference between the digit at the hundred's place and the digit at the unit's place = $9 – 1 = 8$

Hence, option (C) is correct.