Answer : (C) 64
$\prod_{k=1}^{5} k! = 1! *2!*3!*4!*5!$
$ = 1*2*1*3*2*1*4*3*2*1*5*3*2*1 = 2^8 *3^3 *5$
$\text{Total number of divisors } = (8+1)*(3+1)*(1+1) = 72$
$\text{total number of odd divisors} = (3+1)*(1+1) = 8$
$\text{total number of even divisors} = \text{total number of divisors} – \text{total number of odd divisors}$
$72 – 8 = 64$