Let the boxes be marked as $A, B$ and $C$. We have to ensure that no box remains empty and all five balls have to put in. There will be two possibilities :
(i) Any two box containing one ball each and $3$ rd box containing $3$ balls. Number of ways
$ = A (1) \,B(1)\,C(3) =\, ^{5}C_{1} \cdot\, ^{4}C_{1} \cdot\,^{3}C_{3} $
$= 5 \cdot 4 \cdot 1$
$ = 20$
(ii) Any two box containing $2$ balls each and third containing $1$ ball, the number of ways
$ = A(2)\, B(2)\, C(1) =\, ^{5}C_{2} \cdot\, ^{3}C_{2} \cdot\, ^{1}C_{1}$
$ = 10 \times 3 \times 1 $
$= 30$
Since, the box containing $1$ ball could be any of the three boxes $A, B, C$.
Hence, the required number of ways
$= 30 \times 3 = 90$.
Hence, total number of ways $= 60 + 90$
$ = 150$.