We Only Know Two Of Each!

The first three boxes contain each shape(Circle, square or triangle) twice in them.

Hence, if we add the weight of the first 3 boxes, then the result will be twice the total weight of the 3 shapes.

Doing so, $13+18+17=48$

Therefore, $\dfrac{48}{2}=24$ is the required number.

$\begin{aligned} \text{Square}+\text{Triangle}&=13\qquad(1)\\ \text{Circle}+\text{Square}&=18\qquad(2)\\ \text{Circle}+\text{Triangle}&=17\qquad(3)\\ (1)+(2)+(3)\implies 2(\text{Circle}+\text{Square}+\text{Triangle})&=48\\ \boxed{\text{Circle}+\text{Square}+\text{Triangle}=24} \end{aligned}$

$S$ = square

$T$ = triangle

$C$ = circle

We have three equations: $S+T=13; C+S=18$ and $C+T=17$

Adding the three equations, we get

$2(S+C+T)=48$

It follows that,

$S+C+T=$ $\color{#D61F06}\large \boxed{24}$