Given,

Mass of solute $\left(w_B\right)=10 \mathrm{~g}$
Molar mass of solute $\left(M_B\right)=M_B$
Mass of solvent $\left(w_A\right)=360 \mathrm{~g}$
Relative lowering in vapour pressure of solution
$$
=5 \times 10^{-3}
$$

Molar mass of water $\left(M_A\right)=18 \mathrm{~g} \mathrm{~mol}^{-1}$
$\because \frac{\Delta p}{p^{\circ}}=$ Relative lower in vapour-pressure of solution.
$$
\frac{\Delta p}{p^{\circ}}=\chi_B \Rightarrow \frac{n_B}{n_A+n_B}=5 \times 10^{-3}
$$
where, $n_A$ and $n_B$ are number of moles of solvent $(A)$ and solute $(B)$ respectively.
$$
\begin{aligned}
n_A & =\frac{360}{18}=20 \\
n_B & =\frac{w_B}{M_B}=\frac{10}{M_B} \\
\because \quad 5 \times 10^{-3} & =\chi_B=\frac{\frac{10}{M_B}}{20+\frac{10}{M_B}}
\end{aligned}
$$
$$
\begin{aligned}
& 5 \times 10^{-3}=\frac{\frac{10}{M_B}}{\frac{20 M_B+10}{M_B}} \\
& 5 \times 10^{-3}=\frac{10}{20 M_B+10} \\
&\left(20 M_B+10\right) 5 \times 10^{-3}= 10 \\
& \text { or, } \quad 20 M_B+10=\frac{10}{5} \times 10^3 \\
& 20 M_B+10=2000 \Rightarrow 20 M_B=1990 \\
& M_B=\frac{1990}{20}=99.5 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$

Hence, option (b) is the correct answer.