The moment of inertia \(I\) of a point mass \(m\) about an axis passing through it perpendicular to its plane is given by the formula:

\[ I = m \cdot r^2 \]

where \( r \) is the distance of the point mass from the axis.

In this problem, we have four point masses, each of mass \(1 \, \text{kg}\), placed at the vertices of a square of side \(2 \, \text{m}\). The distance of each mass from the axis passing through one of the vertices perpendicular to the plane of the square is the diagonal of the square, which is \( \sqrt{2} \times 2 \) or \( 2\sqrt{2} \) meters.

So, the moment of inertia \( I \) for each point mass is:

\[ I_{\text{each}} = 1 \, \text{kg} \times (2\sqrt{2})^2 = 8 \, \text{kg m}^2 \]

Since there are four identical masses, the total moment of inertia about the given axis is:

\[ I = 4 \times I_{\text{each}} = 4 \times 8 = 32 \, \text{kg m}^2 \]

Therefore, the correct answer is option (a) \( I = 32 \, \text{kg m}^2 \).