A refrigerator's coefficient of performance is expressed as $\mathrm{COP}=\frac{\mathrm{Useful} \mathrm{Cooling}}{\mathrm{Work} \mathrm{Required}}$

$=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}$.

Given here $\mathrm{COP}=\frac{1}{3}$, $Q_1=200 \mathrm{J}$.

Substituting in above relation, we have,

$\frac{1}{3}=\frac{Q_2}{200-Q_2}$

$200-Q_2=3Q_2\Rightarrow 4Q_2=200$

$\Rightarrow Q_2=50 \mathrm{J}$.

Now, the work done on the working substance is $W=Q_1-Q_2=200-50=150 \mathrm{J}$.