Step 1: Energy balance
$\Delta U=Q-W=0$ for an isothermal ideal gas.
Step 2: Heat equals work
$Q=W=-150\ \text{J}$.
Step 3: Eliminate
Magnitude 150 J, not 300 J, and the sign is negative, so heat is removed. Option (D) is wrong because heat flow must equal the work. Option (B) is correct.
Final Answer:
150 J of heat leaves the gas.
\[ \boxed{\text{(B) }\text{150 J of heat has been removed from the gas}} \]