Answer
About $136g/mol$.
Work Step by Step
A monoprotic acid, by definition, only releases 1 proton ($H^{+}$) per molecule to the solution in a reaction with a base (e.g: $HCl$, $HNO_{3}$, $CH_{3}COOH$,...).
In this case, this organic acid could be written as $HC_{x}H_{y-1}O_{z}$.
Because $NaOH$ also only releases 1 hydroxide ion ($OH^{-}$) per molecule, their ratio in this acid-base reaction is 1:1.
$NaOH+HC_{x}H_{y-1}O_{z}->C_{x}H_{y-1}O_{z}Na+H_{2}O$
$n_{NaOH}=C*V=0.1008M*0.015L=1.512*10^{-3}(mol)$
$->n_{HC_{x}H_{y-1}O_{z}}=1.512*10^{-3}(mol)$
Molar mass of the acid: $M_{HC_{x}H_{y-1}O_{z}}=\frac{n}{m}=\frac{0.2053g}{1.512*10^{-3}mol}=135.78(g/mol)$ rounded to: $136g/mol$
