$(10a+b)\times(10c +b)=100d+10d+d$

$100ac+10(ab+bc)+b^2=100d+10d+d$

$ac=d\ldots\ldots (1)$

$b(a+c)=d\ldots\ldots (2)$

$b^2=d\ldots\ldots (3)$

Two Cases : $b=2,d=4$ or $b=3,d=9$

Case 1 :

From (2) : $a+c = 2 \rightarrow $ not possible

Case 2 :

From (2) : $a+c= 3$

then $ac$ must be 2 $(1)$ is not satisfied. No possible cases.