if 0.15 g of a solute is dissolved in 15 g of solvent , is boiled at a temperature higher by 0.216 celcius, than that of pure solvent . the molecular weight of substance ( molal elevation const=2.16 celcius)

Given: Elevation in boiling point, ∆T_b = 0.216⁰C
Mass of solute = 0.15 g
Mass of solvent = 15 g
You have not mention the wrong unit of molal elevation constant. Unit is K Kg mol^-1
Molal Elevation constant = 0.216 K kg mol^-1
​As we know 
∆T_b = K_b m
where m is the molality.

$\text{Molality }= \frac{{\text{w}}_{\text{2}}\text{×1000}}{{\text{M}}_{\text{2}}{\text{×w}}_{\text{1}}}$
where w₂ is the mass of solute, M₂ is the molar mass of solute and w₁ is the mass of solvent.
 
$$\begin{gathered} 0.216 = 0.216\times \frac{0.15\times 1000}{{\text{M}}_{\text{2}}\times 15} \\ {\text{M}}_{\text{2}}{\text{ = 10 g mol}}^{\text{-1}} \end{gathered}$$
Molar mass of solute = 10 g mol^-1