The special number

$0 <a<10$ also $0<b<10$ and $a$ cannot be equal to $b$ .

$b^{3}-a^{2}=10a+b$

$b^{3}-b=10a-a^{2}$

$b(b+1)(b-1)=a(10+a)$

the product of 3 consecutive integers is divisible by 3!=6

this means either 6 divides $a$ or $(10+a)$

Possible values of $a$ are 2,3,6.

since smallest is asked so first taking 2 in the equation we find 3 satisfies the value of $b$ ,therefore the number is $\boxed{23}$ .