Step 1: Convert every quantity into consistent Darcy, cgs units:
In the Darcy unit system, Darcy's law reads $q(\text{cm}^3/\text{s}) = \dfrac{k(\text{darcy}) \, A(\text{cm}^2) \, \Delta p(\text{atm})}{\mu(\text{cP}) \, L(\text{cm})}$. Convert each given quantity: permeability $k = 50\ \text{mD} = 0.05$ darcy. Area $A = 6000\ \text{ft}^2$, and since $1\ \text{ft}^2 = 929.03\ \text{cm}^2$, this gives $A = 6000 \times 929.03 = 5{,}574{,}182\ \text{cm}^2$. Length $L = 2000\ \text{ft}$, and since $1\ \text{ft} = 30.48\ \text{cm}$, this gives $L = 2000 \times 30.48 = 60{,}960\ \text{cm}$. Pressure drop $\Delta p = 300\ \text{psi}$, and since $1\ \text{atm} = 14.696\ \text{psi}$, this gives $\Delta p = 300/14.696 = 20.414\ \text{atm}$. Viscosity stays $\mu = 2\ \text{cP}$.
Step 2: Substitute into the Darcy unit equation:
Numerator: $k \times A \times \Delta p = 0.05 \times 5{,}574{,}182 \times 20.414$. First, $0.05 \times 5{,}574{,}182 = 278{,}709$. Then $278{,}709 \times 20.414 = 5{,}689{,}540\ \text{cm}^3\cdot\text{cP/s}$ approximately. Denominator: $\mu \times L = 2 \times 60{,}960 = 121{,}920$. Dividing gives $q = 5{,}689{,}540 / 121{,}920 = 46.67\ \text{cm}^3/\text{s}$.
Step 3: Convert the flow rate from cm³/s to bbl/day:
First convert to cm3/day: $46.67 \times 86400 = 4{,}032{,}000\ \text{cm}^3/\text{day}$. Since $1\ \text{bbl} = 5.61\ \text{ft}^3$ and $1\ \text{ft}^3 = 28{,}316.8\ \text{cm}^3$, one barrel equals $5.61 \times 28{,}316.8 = 158{,}898\ \text{cm}^3$. Dividing the daily volume by this gives $q = 4{,}032{,}000 / 158{,}898 = 25.38\ \text{bbl/day}$.
Step 4: Compare with the field unit result and round:
This matches the value obtained directly from the field unit form of Darcy's law within rounding, confirming the calculation. Rounding to one decimal place: $q \approx 25.4\ \text{bbl/day}$.
Final Answer:
\[ \boxed{25.4 \text{ bbl/day}} \]