Heat extracted by ice at $0^{\circ }\mathrm{C}$ to melt $=$ Heat given by water to reach $0^{\circ }\mathrm{C}$ from ${40}^{\circ }\mathrm{C}$.
$\mathrm{ML}=\mathit{m}\mathit{c}\mathit{\theta }$
Where, $\mathit{\theta }=$ change in temperature of water
$\mathrm{L}=$ Specific Latent heat of ice
$\mathit{c}=$ Specific heat capacity of water
$\mathit{M}=$ Mass of ice required
$\mathit{m}=$ mass of water
Heat gained by ice $=$ Heat lost by water
$\mathrm{M}\times 336=300\times 4.2\times \left(40-0\right)$
$\mathrm{M}=\frac{300\times 4.2\times 40}{336}=150\mathrm{gm}$
$\therefore$ Ice is required to lower the temperature of water $=150\mathrm{gm}$.