$\frac{\Delta V}{V}=\frac{-\Delta P}{B}$   $\left\{\begin{array}{l}\Delta V=10 \% \text { of } V \text { ( } \because \text { Pressure increases volume must } \\ \text { If } \begin{array}{l}\text { decreases by } 10 \% \text { so we will use a +ve sign) }\end{array} \\ \begin{array}{rl}\Rightarrow \Delta V & =100 cc \\ \Delta P & =P_2-P_1 \\ & =1.165 \times 10^5-1.01 \times 10^5\end{array}\end{array}\right.$

Substituting the values

$\frac{-10}{100}=\frac{-\left(1.165 \times 10^6-1.01 \times 10^6\right)}{B}$

$\frac{1}{10}=\frac{.155 \times 10^5}{B}$

$B=1.55 \times 10^5 \,Pa$