Step 1: Write Bg using the real gas equation: Using the real gas law at reservoir and standard conditions, the gas formation volume factor can be written as $B_g = \dfrac{P_{sc} z T}{T_{sc} P}$, where $P_{sc}$ and $T_{sc}$ are standard pressure and temperature, $z$ is the compressibility factor, and $P$, $T$ are reservoir pressure and temperature.
Step 2: Write E as the inverse quantity: The gas expansion factor is defined as the number of standard volumes obtained from one reservoir volume, so by definition $E = \dfrac{1}{B_g} = \dfrac{T_{sc} P}{P_{sc} z T}$.
Step 3: Interpret the formula: This formula shows E rises when reservoir pressure P rises or when z falls, and Bg does the exact opposite, it rises when P falls or z rises. The two quantities always move in opposite directions for the same gas.
Step 4: Test with a numeric idea: If a reservoir volume of gas is highly compressed at high pressure, one reservoir barrel yields a large number of standard cubic feet at surface, so E is large and Bg, being its reciprocal, is small. This numeric check again confirms the inverse relation.
Step 5: Match with the given options: Only option A, $E \propto \dfrac{1}{B_g}$, agrees with this inverse relationship, while the direct and higher power relationships in options B, C and D do not follow from either the definitions or the real gas equation.
Final Answer: \[ oxed{E \propto \dfrac{1}{B_g}} \]