A

Step 1: Analyze the circuit and identify the components. There are five resistors of 200 ohms each and two batteries of 3 volts.
Step 2: Calculate the total resistance in the open circuit condition. Since two resistors are in parallel and in series with others, we have to calculate accordingly. The two in parallel will have an equivalent resistance $R_{parallel} = \frac{R \cdot R}{R + R} = \frac{200 \cdot 200}{200 + 200} = 100 \text{ ohms}$.
Step 3: Total resistance in series = 100 ohms + two series resistors = 100 + 200 + 200 = 500 ohms.
Step 4: Using Ohm’s law for open circuit voltage: $V = I \cdot R = \frac{V_{total}}{R_{total}} \cdot R_{open} = 3 ext{ V} \cdot 2 = 6 ext{ volts}$.
Step 5: For short-circuit current, total resistance is 0. Therefore, $I = \frac{E}{R} = \frac{3 + 3}{0} \to \text{undefined/infinitely large current}$. Hence, the short circuit current is determined by the internal resistance. Assuming it negligible, in practical it may be limited to (3/200) + (3/200) = 0.03 A.
Therefore, V(open circuit) = 6 V and I(short circuit) = 0.03 A.