jmc

First, we use the property $a\equiv b(\mathrm{mod}m)$ implies $ac\equiv bc(\mathrm{mod}m)$.

Since all numbers with units digit $3$ have a remainder of $3$ when divided by $5$ and there are $20$ numbers, $$3\times 13\times 23\times 33\times \dots \times 183\times 193\equiv 3^{20}(\mathrm{mod}5).$$Next, we also use the property $a\equiv b(\mathrm{mod}m)$ implies $a^c\equiv b^c(\mathrm{mod}m)$.

Since $3^4\equiv 81\equiv 1(\mathrm{mod}5)$, and $3^{20}=(3^4{)}^5$, then $3^{20}\equiv 1^5\equiv \boxed{1}(\mathrm{mod}5)$.