Let the two consecutive natural numbers which are multiples of 3 be 3x and 3(x+1)
Now, 3x(3x+3)=810
⇒x2+x=90
⇒x2+x−90=0
⇒(x+10)(x−9)=0
⇒x=9 or x=−10
Rejecting negative value of x, because number are natural. We have x=9
Hence, the required numbers are 27 and 30