jmc

An integer congruent to $3(\mathrm{mod}11)$ can be written as $11n+3$. Therefore, we have the inequality $$-100\le 11n+3\le 100$$We solve for the inequality by subtracting each term by $3$ and then dividing by $11$ to get $$-103\le 11n\le 97\to -\frac{103}{11}\le n\le \frac{97}{11}$$The smallest integer greater than $-\frac{103}{11}$ is $-9$ and the largest integer less than $\frac{97}{11}$ is $8$. There are $\boxed{18\text{ integers}}$ from $-9$ to $8$.