A) The above figure is the label diagram of the jewelry box.
B) The equation of the volume will be,
![\begin{gathered} \text{Volume of a cuboid=length}* width* height \\ \text{length}=(x+8) \\ \text{width}=x \\ \text{height}=12\text{inches} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6ddq9dhdkmpxjq6vqeec372ukzn1mb2qs9.png)
![\begin{gathered} \text{Volume}=(x+8)(x)(12)in^3 \\ =12(x^2+8x)=(12x^2+96x)in^3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/aqp3lm9tvtt52v74p5mkoqbclqpfv3jse9.png)
The equation for the volume of the box is (12x²+96x)in³.
C) To solve for the width of the box in metres, we will equate the equation of the volume to 240in³,
![\begin{gathered} 12x^2+96x^{}=240 \\ \text{collect like terms} \\ 12x^2_{}+96x-240=0 \\ \text{Divide through by 12} \\ (12x^2)/(12)+(96x)/(12)-(240)/(12)=0 \\ x^2+8x-20=0 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jkm4n3liv59at8c68qn0tcmbs2rr4q2awz.png)
![\begin{gathered} x^2+10x-2x-20=0 \\ x(x+10)-2(x+10)=0 \\ (x-2)(x+10)=0 \\ x-2=0\text{ or x+10=0} \\ x=2\text{ or x=-10} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7ltogkdieuanlddx1zt05ypqh3qnz007az.png)
The width can never be negative, therefore the width is 2in.
Finally let us now convert the width to meters,
![\begin{gathered} 1\text{inch}=0.0254\text{metres} \\ 2\text{inches}=2*0.0254metres \\ =0.0508metres \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/se4bs8zvx78z211j8ufl58ks91hpgly4qd.png)
Hence, the width is 0.0508metres.