jmc

An integer congruent to $7(\mathrm{mod}19)$ can be written as $19n+7$. Therefore, we have the inequality $$100\le 19n+7\le 999.$$We solve for the inequality by subtracting each term by $7$ and then dividing by $19$ to get $$93\le 19n\le 992⟹\frac{93}{19}\le n\le \frac{992}{19}.$$The smallest integer greater than $\frac{93}{19}$ is $5$ and the largest integer less than $\frac{992}{19}$ is $52$. There are $52-4=\boxed{48}$ integers from $5$ to $52$ inclusive.