Here, mass of metal, m = 0.20 kg =200 g
Fall in temperature of metal

           $\mathrm{\Delta }T={150}^{\circ }\mathrm{C}-{40}^{\circ }\mathrm{C}={110}^{\circ }\mathrm{C}$

Ife is specific heat of the metal, then heat lost by the metal,

                     $\mathrm{\Delta Q}=\mathrm{mc\Delta T}=200\mathrm{s}\times 110$   ……….(i)

Volume of water = 150 cc
Mass of water, m' = 150g
Water equivalent of calorimeter

                         $\mathrm{w}=0.025\mathrm{kg}=25\mathrm{g}$

Rise in temperature of water in calorimeter

${\mathrm{\Delta T}}^{\mathrm{\prime }}={40}^{\circ }\mathrm{C}-{27}^{\circ }\mathrm{C}={13}^{\circ }\mathrm{C}$

Heat gained by water and calorimeter

$\begin{array}{l} {\mathrm{\Delta Q}}^{\mathrm{\prime }}=\left({\mathrm{m}}^{\mathrm{\prime }}+\mathrm{w}\right){\mathrm{\Delta T}}^{\mathrm{\prime }}=(150+25)\times 13\\ {\mathrm{\Delta Q}}^{\mathrm{\prime }}=175\times 13 .........\left(\mathrm{ii}\right)\\ \mathrm{As} \mathrm{\Delta Q}={\mathrm{\Delta Q}}^{\mathrm{\prime }}\end{array}$

$\therefore$ From Eqs. (i) and (i), we get

$\begin{array}{c}200\times \mathrm{s}\times 110=175\times 13\\ \mathrm{s}=\frac{175\times 13}{200\times 110}\approx 0.10 \mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C}\end{array}$