The sum of all the proper divisors of 9900 is

$9900=9\times 11\times 100=2^2\cdot 5^2\cdot 3^2\cdot {11}^1$

The number of divisors will be 

$\begin{array}{l}(2+1)(2+1)(2+1)(1+1)\\ =3\times 3\times 3\times 2=54\end{array}$

Exclude 1 and 9900 the numbers itself and hence proper divisors are 54- 2 = 5?. Sum of all the divisors is 

$\begin{array}{l}\left(2^0+2^1+2^2\right)\left(5^0+5^1+5^2\right)\left(3^0+3^1+3^2\right)\left({11}^0+{11}^1\right)\\ =7\times 31\times 13\times 12=33852\end{array}$

This also includes 1 and 9900 which are not proper. Hence the sum of proper divisors is 

33852 - 9900 - 1 = 23951.