$Sum of all three digit numbers (no digit being zero)$

$having the property that all digits are perfect squares,$

$\begin{array}{l}\text{The non zero perfect square digits re 1,4, 9}\\ \text{1 can occur at units place 3}\cdot \text{3=9 ways}\\ \text{Sum due to 1 at unit place is 1}\cdot \text{9=9 ways}\\ \text{Sum due to 1 at tens place is 1}\cdot \text{10}\cdot \text{9=90 ways}\\ \text{sum due to 1 at hundereds place 1}\cdot \text{100}\cdot \text{9=900\hspace{0.17em}ways}\\ \text{4 and 9 in a similar way}\\ \text{Thus the sum of the desired number is }\left(1+4+9\right)\left(1+10+100\right)9=13986\end{array}$