Answer:
Approximately [tex]1.5\; \rm kg[/tex]. (Assuming that this table is level, and that the gravitational field strength is [tex]g = 9.8\; \rm N \cdot kg^{-1}[/tex].
Explanation:
Convert the dimensions of this book to standard units:
[tex]\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m[/tex].
[tex]\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.25\; \rm m[/tex].
Calculate the surface area of this book:
[tex]0.30\; \rm m \times 0.25\; \rm m = 0.075\; \rm m^{2}[/tex].
Pressure is the ratio between normal force and the area over which this force is applied.
[tex]\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}[/tex].
Equivalently:
[tex](\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area})[/tex].
In this question, [tex]\text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}[/tex].
It was found that [tex](\text{contact Area}) = 0.075\; \rm m^{2}[/tex] (assuming that the entire side of this book is in contact with the table.
Hence:
[tex]\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}[/tex].
If that the table is level, this normal force would be equal to the weight of this book:
[tex]\text{weight} = 15\; \rm N[/tex].
Assuming that the gravitational field strength is [tex]g = 9.8\; \rm N \cdot kg^{-1}[/tex]. The mass of this book would be:
[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}[/tex].
