Answer: D

Solution:

The amounts $30,31,32,33,34,35$ are all attainable (for instance $30=6\cdot 5,$ $31=6+10+15,$ $32=6\cdot 2+10\cdot 2,$ $33=6\cdot 3+15,$ $34=6\cdot 4+10,$ $35=10+10+15$). Adding $6$'s then reaches every larger amount. Checking below, $29$ is impossible: without a $15$-coin a sum of $6$'s and $10$'s is even, while with one $15$-coin the remaining $14$ cannot be written as $6a+10b;$ two $15$-coins already exceed $29.$ So the largest impossible amount is $29,$ whose digit sum is $2+9=11.$

Thus, the correct answer is D.