Let $d(n)$ be the number of positive integers that divide the integer $n$. For all positive integral divisors $k$ of $64800$, what is the sum of numbers $d(k)$?

$ 
\textbf{(A)}\ 1440
\qquad\textbf{(B)}\ 1650
\qquad\textbf{(C)}\ 1890
\qquad\textbf{(D)}\ 2010
\qquad\textbf{(E)}\ \text{None of above}
$
1. First, we need to find the prime factorization of \( 64800 \). We can start by breaking it down step-by-step:
   \[
   64800 = 648 \times 100
   \]
   \[
   648 = 2^3 \times 81 = 2^3 \times 3^4
   \]
   \[
   100 = 2^2 \times 5^2
   \]
   Therefore,
   \[
   64800 = 2^3 \times 3^4 \times 2^2 \times 5^2 = 2^5 \times 3^4 \times 5^2
   \]

2. Next, we need to find the sum of \( d(k) \) for all positive integral divisors \( k \) of \( 64800 \). The function \( d(n) \) is multiplicative, meaning \( d(n) \) for \( n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \) is given by:
   \[
   d(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   For \( 64800 = 2^5 \times 3^4 \times 5^2 \), we have:
   \[
   d(64800) = (5+1)(4+1)(2+1) = 6 \times 5 \times 3 = 90
   \]

3. To find the sum of \( d(k) \) for all divisors \( k \) of \( 64800 \), we use the fact that the sum of \( d(k) \) for all divisors \( k \) of \( n \) is given by:
   \[
   \sum_{k|n} d(k) = \prod_{i=1}^m \left( \sum_{j=0}^{e_i} d(p_i^j) \right)
   \]
   where \( n = p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m} \).

4. For \( 64800 = 2^5 \times 3^4 \times 5^2 \), we calculate:
   \[
   \sum_{j=0}^{5} d(2^j) = d(1) + d(2) + d(2^2) + d(2^3) + d(2^4) + d(2^5) = 1 + 2 + 3 + 4 + 5 + 6 = \frac{6(7)}{2} = 21
   \]
   \[
   \sum_{j=0}^{4} d(3^j) = d(1) + d(3) + d(3^2) + d(3^3) + d(3^4) = 1 + 2 + 3 + 4 + 5 = \frac{5(6)}{2} = 15
   \]
   \[
   \sum_{j=0}^{2} d(5^j) = d(1) + d(5) + d(5^2) = 1 + 2 + 3 = \frac{3(4)}{2} = 6
   \]

5. Therefore, the sum of \( d(k) \) for all divisors \( k \) of \( 64800 \) is:
   \[
   \sum_{k|64800} d(k) = 21 \times 15 \times 6 = 1890
   \]

The final answer is \(\boxed{1890}\)