What is the question answer........

Solution,
Given,
volume of milk = 250 mL
volume of water = 100 mL
Temperature of milk = 80^oC
Temperature of water = 40^oC
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Density of water = 1.00 gm/mL
Density of milk = 1.027 gm/mL
So, mass of milk, $m_1= 1.027\times 250=256.75 gm$
mass of water, $m_2= 1.0\times 100=100 gm$
The specific heat capacity of milk is 3.93 Jg^-1oC^-1 and that of water is 4.2 Jg^-1oC^-1.
Let the final temperature of the mixture be T.
Heat lost by milk = heat gained by water
$$\begin{gathered} m_1{c}_1\left(80-T\right)=m_2{c}_2\left(T+40\right) \\ 257\times 3.93\times \left(80-T\right)=100\times 4.2\times \left(T+40\right) \\ \frac{\left(80-T\right)}{\left(T+40\right)}=\frac{100\times 4.2}{257\times 3.93}=0.416 \\ 80-T=0.416T+16.64 \\ 63.36=1.416T \\ T=\frac{63.36}{0.416}=44.74 = {45}^{o}C \end{gathered}$$
Thus, CORRECT option is (D)