There are positive integers that have these properties:

the sum of the squares of their digits is 50, and
each digit is larger than the one to its left.
The product of the digits of the largest integer with both properties is
$\text{(A)}\ 7 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 60$
Five-digit numbers will have a minimum of $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$ as the sum of their squares if the five digits are distinct and non-zero.  If there is a zero, it will be forced to the left by rule #2.
No digit will be greater than $7$, as $8^2 = 64$.
Trying four digit numbers $WXYZ$, we have $w^2 + x^2 + y^2 + z^2 = 50$ with $0 < w < x < y < z < 8$
$z=7$ will not work, since the other digits must be at least $1^2 + 2^2 + 3^2 = 14$, and the sum of the squares would be over $50$.
$z=6$ will give $w^2 + x^2 + y^2 = 14$.  $(w,x,y) = (1,2,3)$ will work, giving the number $1236$.  No other number with $z=6$ will work, as $w, x,$ and $y$ would have to be greater. 
$z=5$ will give $w^2 + x^2 + y^2 = 25$.  $y=4$ forces $x=3$ and $w=0$, which has a leading zero, and then we have 345 which is a 3-digit number. 
$z=4$ can only give the number $1234$, which does not satisfy the condition of the problem.
Thus, the number in question is $1236$, and the product of the digits is $36$, giving $\boxed{C}$ as the answer.