Out of the given numbers,the numbers which are multiple of 5 are 5,10,15,20.....145
This is an AP whose first term a=5
Common difference =5
$n^{th}$ term,$t_n=145$
$145=5+(n-1)\times 5$
$\Rightarrow n=29$
Similarly if total number of numbers is $m$ then
$151=1+(m-1)\times 2$
$m-1=\large\frac{151-1}{2}$
$m=76$
So number of ways in which product is a multiple of 5
$\Rightarrow$ Both two numbers from 5,15,25....145+one number from 5,15,25,....145 and one from remaining numbers
$\Rightarrow 29C_2+29C_1\times 76C_1$
$\Rightarrow 29\times 14+29\times 76$
$\Rightarrow 29\times 90$
Hence (D) is the correct answer.