jmc

There are $12$ months in a year. January, March, May, July, August, October, and December have $31$ days, so there are $7$ months with $2$ days that have a $3$ in the tens digit. April, June, September, and November each have $1$ day with a $3$ in the tens place, and February has none. Therefore, there are $7\times 2+4=\boxed{18}$ days with a $3$ in the tens digit.

Alternatively, this problem can be solved by subtraction. There are $365$ days in a year, and $11$ months have $29$ days without a $3$ in the tens digit, and the last month has $28$ (assuming a non-leap year). $365-(11\times 29)-28=\boxed{18}$