To find the gate voltage \( V_G \) of the n-channel MOSFET, we start by analyzing the circuit. The MOSFET is in saturation since it's connected as a source follower with \( V_{DD} > V_{TH} \). The current equation in saturation is:

\[ I_D = \frac{1}{2}\mu C_{ox}\left(\frac{W}{L}\right)(V_{GS} - V_{TH})^2 \]

Here, the drain current (\( I_D \)) through the 1 kΩ resistor can be calculated as:

\[ I_D = \frac{V_{DD} - V_G}{1 \, \text{k}\Omega} \]

Equating both expressions for \( I_D \):

\[\frac{V_{DD} - V_G}{1 \, \text{k}\Omega} = \frac{1}{2} \cdot 2 \, \text{mA/V}^2 \cdot (V_G - V_{TH})^2\]

\[V_{DD} - V_G = (V_G - V_{TH})^2\]

Substitute \( V_{DD} = 5 \, \text{V} \) and \( V_{TH} = 1 \, \text{V} \):

\[5 - V_G = (V_G - 1)^2\]

Expanding and simplifying:

\[5 - V_G = V_G^2 - 2V_G + 1\]

\[V_G^2 - V_G - 4 = 0\]

Using the quadratic formula \( V_G = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), \( c = -4 \):

\[V_G = \frac{1 \pm \sqrt{1 + 16}}{2}\]

\[V_G = \frac{1 \pm \sqrt{17}}{2}\]

Since \( \sqrt{17} \approx 4.12 \):

\[V_G = \frac{1 \pm 4.12}{2}\]

Calculating the positive root (ensures \( V_{GS} > V_{TH} \)):

\[V_G = \frac{5.12}{2} = 2.56 \, \text{V}\]

Thus, the computed \( V_G \) of 2.56 V falls within the expected range of 2.5 to 2.6 V.