The percentage composition by weight of an aqueous solution of a solute (molecule mass150) which boils at 373.26 is (K_b = 0.52)

The solute content in the solution directly relates to the increase in boiling point (Tb).It can be calculated using the equation below. 

$∆$T_b=K_b-m, where $∆$T_b is 0.2 (elevation in B.P), K_b=0.52(constant) and m is molality.

m=$∆$T_b/K_b=0.26/0.52 =0.5mole/kg

 In 1000 g of solvent, there will be 0.5 moles of solute in 1000 g of water, which is the definition of molality. 

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\times \frac{{\mathrm{W}}_{\mathrm{s}}}{{\mathrm{M}}_{\mathrm{s}}}\times \frac{1000}{{\mathrm{W}}_{\mathrm{o}}} \\ \frac{∆{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{b}}}=0.5=\frac{{\mathrm{W}}_{\mathrm{s}}}{150}\times \frac{1000}{1000} \end{gathered}$$

thus w_s=75

percentage composition=  $\frac{75\times 100}{1000}=7.5\%$

Hence, the correct option (C).