We have $10$ teams.
First team can play with any other team in $9$ ways.
Similarly second can play with any other in $8$ ways (excluding the first team), $3^{rd}$ team can play $7$ games and so on.
So, total number of games possible $=9+8+7\ldots +1=\dfrac{n(n+1)}{2} =\dfrac{10\times 9}{2}=45$
Since, each team plays $2$ games against each other, total games $=45\times 2=90$
Correct Answer: $D$