The efficiency of the heat engine is

Q1=150j

Q2=100j

STEP 1: Given is the following heat engine

Given, $Q_s = 100\;J$

$W = 50\;J$

where, $Q_r =$ heat rejection

$w =$ work done

$Q_s =$ heat supplied

$Q_1 = w + Q_2$

=> $Q_1 = 50 + 100$

=> $Q_s = Q_1 = 150\;J$

STEP 2: Calculate efficiency

=> efficiency $\eta = 1 - \dfrac{Q_2}{Q_1}$

$= 1 - \dfrac{100}{150}$

$= \dfrac{150-100}{150}$

$= \dfrac {50}{150} \times 100 \% = 33.33\%$

Thus, option B is correct.