Just 3 integers...

$\begin {matrix} 100^{99}-100^{98} & < & 100^{99} - 99^{98} & < & 100^{99} - 100 ^{97}\\ 10^{198}-10^{196} & < & 100^{99} - 99^{98} & < & 10^{198} - 10^{194}\\ 10^{196}(100-1) & < & 100^{99} - 99^{98} & < & 10^{194}(10000 -1)\\ 99\times 10^{196} &< &100^{99} - 99^{98} & < & 9999\times 10^{194}\\ 9.9\times 10^{197} &< & 100^{99} - 99^{98} & < & 9.999\times 10^{197} \end {matrix}$

Therefore, $\space 100^{99} - 99^{98} \space$ has $198$ digits.

Find the # of digits in $100^{99} - 99^{98}$ .

This expression is equivalent to $\frac{100^{98}}{100^{98}} (100^{99} - 99^{98})$ .

Dividing through, we get: $100^{98} \left( 100 - \left(\frac{99}{100} \right)^{98} \right)$

Since $0 \lt \frac{99}{100} \lt 1$ , hence $0 \lt \left( \frac{99}{100} \right) ^{98} \lt 1$ ,

So $99 \lt \left[ 100 - \left( \frac{99}{100} \right) ^{98} \right] \lt 100$ . and this number has 2 digits in the integer-part.

Now your resultant number is in this form: $100^{98} \left( \mbox{99...} \right)$

$100^{98}$ , which has $196$ zeroes after the $1$ , when multiplied with $99...$ will have $196$ places after the $99$ , which gives a total of $\boxed{198}$ digits.