To find the output voltage \( V_o \) in the given circuit, follow these steps. The circuit includes an ideal op-amp and a MOSFET operating in the saturation region.

1. Since the op-amp is ideal, the voltage at the inverting and non-inverting terminals will be equal due to the virtual short concept. Thus, V^+ = V^- = 2 \, \text{V}.
2. The voltage at the inverting terminal is governed by the voltage divider formed by the 3V source and the 4kΩ resistor. V^- = 2 \, \text{V} is set by the circuit configuration.
3. In the saturation region, the drain current \( I_D \) of the MOSFET is relatively constant and is determined by the gate voltage, derived from the op-amp output.
4. Assuming the MOSFET is perfectly controlled in saturation, the voltage V_o = I_D \times 1\text{k}\Omega is determined by the current through the 1kΩ resistor.
5. Since the gate-source voltage is set by the op-amp to achieve the virtual short condition, and \( V^- = 2 \, \text{V} \), it indirectly sets \( I_D \) such that I_D \times 1\text{k}\Omega = 0.25 \, \text{V}.
6. Thus, the output voltage V_o = 0.25 \, \text{V}.

Therefore, the correct answer is 0.25 V.