Solution:

Prime factorize $18=2\cdot3^2,$ $45=3^2\cdot5,$ and $180=2^2\cdot3^2\cdot5.$

The lcm condition forces $n$ to contain $2^2$ and $5$, while allowing at most $3^2$. The gcd condition $\gcd(n,45)=15=3\cdot5$ forces the exponent of $3$ in $n$ to be exactly $1$ and the exponent of $5$ to be at least $1$.

Therefore $n=2^2\cdot3\cdot5=60,$ and the sum of its digits is $6.$

Thus, B is the correct answer.