jmc

Expanding the binomial coefficient, we get $\left(\genfrac{}{}{0ex}{}{200}{100}\right)=\frac{200!}{100!100!}$. Let the required prime be $p$; then $10\le p<100$. If $p>50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$. The largest such prime is $\boxed{61}$, which is our answer.