To determine the node voltages \( V_1 \) and \( V_2 \), we can use the nodal analysis method. Let's apply Kirchhoff's current law (KCL) at nodes \( V_1 \) and \( V_2 \).

Step 1: Apply KCL at Node \( V_1 \)

Consider the current entering and leaving the node:

\[ 2 = \frac{V_1}{7} + \frac{V_1 - V_2}{15} + \frac{V_1}{3} \]

Multiply through by 105 (LCM of 7, 15, and 3) to clear fractions:

\[ 210 = 15V_1 + 7(V_1 - V_2) + 35V_1 \]

Simplify:

\[ 210 = 15V_1 + 7V_1 - 7V_2 + 35V_1 \]

Combine like terms:

\[ 210 = 57V_1 - 7V_2 \]

Step 2: Apply KCL at Node \( V_2 \)

Consider the current entering and leaving the node:

\[ \frac{V_2 - V_1}{15} + \frac{V_2}{5} = 4 \]

Multiply through by 15 (LCM of 15 and 5) to clear fractions:

\[ V_2 - V_1 + 3V_2 = 60 \]

Simplify:

\[ 4V_2 - V_1 = 60 \]

Step 3: Solve the System of Equations

We now have two equations:

\[ 57V_1 - 7V_2 = 210 \] \[ 4V_2 - V_1 = 60 \]

Rearrange the second equation:

\[ V_1 = 4V_2 - 60 \]

Substitute \( V_1 \) in the first equation:

\[ 57(4V_2 - 60) - 7V_2 = 210 \]

Expand and simplify:

\[ 228V_2 - 3420 - 7V_2 = 210 \] \[ 221V_2 = 3630 \]

Solve for \( V_2 \):

\[ V_2 = 20 \, V \]

Substitute \( V_2 \) back to find \( V_1 \):

\[ V_1 = 4 \times 20 - 60 = 80 - 60 = 20 \, V \]

Thus, the node voltages are \( V_1 = 20 \, V \) and \( V_2 = 20 \, V \).

Conclusion: The correct answer is \( V_1 = 20 \, V \), \( V_2 = 20 \, V \).