Because acceleration was constant throughout the slide, we can show the slide lasted
[tex]a=\dfrac{\Delta v}{\Delta t}\iff-5.6\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{7.4\,\frac{\mathrm m}{\mathrm s}-8\,\frac{\mathrm m}{\mathrm s}}{\Delta t}[/tex]
[tex]\implies\Delta t=0.107\,\mathrm s[/tex]
Also because accleration was constant, we know the average velocity of the book was
[tex]\bar v=\dfrac{7.4\,\frac{\mathrm m}{\mathrm s}+8\,\frac{\mathrm m}{\mathrm s}}2=7.7\,\dfrac{\mathrm m}{\mathrm s}[/tex]
Average velocity is also given by
[tex]\bar v=\dfrac{\Delta x}{\Delta t}\iff7.7\,\dfrac{\mathrm m}{\mathrm s}=\dfrac{\Delta x}{0.107\,\mathrm s}[/tex]
so the width of the desk must have been
[tex]\Delta x=0.824\,\mathrm m[/tex]
which means C is the most likely answer.