If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.
Doing some casework:
If the middle digit is $1$, possible numbers range from $111$ to $210$. So there are $2$ numbers in this case.
If the middle digit is $2$, possible numbers range from $123$ to $420$. So there are $4$ numbers in this case.
If the middle digit is $3$, possible numbers range from $135$ to $630$. So there are $6$ numbers in this case.
If the middle digit is $4$, possible numbers range from $147$ to $840$. So there are $8$ numbers in this case.
If the middle digit is $5$, possible numbers range from $159$ to $951$. So there are $9$ numbers in this case.
If the middle digit is $6$, possible numbers range from $369$ to $963$. So there are $7$ numbers in this case.
If the middle digit is $7$, possible numbers range from $579$ to $975$. So there are $5$ numbers in this case.
If the middle digit is $8$, possible numbers range from $789$ to $987$. So there are $3$ numbers in this case.
If the middle digit is $9$, the only possible number is $999$. So there is $1$ number in this case.
So the total number of three-digit numbers that satisfy the property is $2+4+6+8+9+7+5+3+1=\boxed{\textbf{(E) }45}$