We can determine the time period of the oscillations by considering that when the cube is depressed by a small displacement, the additional buoyant force provided by the displaced water acts as a restoring force. Here’s a step‐by‐step explanation:

Define the given values:

Side length of the cube, $$L = 10\text{ cm} = 0.1\text{ m}$$.

Mass of the cube, $$m = 10\text{ g} = 0.01\text{ kg}$$.

Density of water, $$\rho = 10^3\ \text{kg/m}^3$$.

Acceleration due to gravity, $$g = 10\ \text{m/s}^2$$.

Cross-sectional area of the cube (face area),

$$A = L^2 = (0.1)^2 = 0.01\text{ m}^2.$$

When the cube is depressed by a small distance $$x$$, the additional volume of water displaced is

$$\Delta V = A\,x.$$

Thus, the additional buoyant force is:

$$F_b = \rho\,g\,\Delta V = \rho\,g\,A\,x.$$

This force acts in the upward (restoring) direction. Notice that the force is proportional to the displacement $$x$$, which is the hallmark of simple harmonic motion (SHM).

The effective spring constant, $$k$$, associated with this SHM is:

$$k = \rho\,g\,A.$$

The period of oscillation for SHM is given by:

$$T = 2\pi \sqrt{\frac{m}{k}},$$

which, upon substituting for $$k$$, becomes:

$$T = 2\pi \sqrt{\frac{m}{\rho\,g\,A}}.$$

Substitute the given values into the expression:

$$T = 2\pi \sqrt{\frac{0.01}{1000 \times 10 \times 0.01}}.$$

Calculate the denominator:

$$1000 \times 10 \times 0.01 = 100,$$

so,

$$T = 2\pi \sqrt{\frac{0.01}{100}} = 2\pi \sqrt{0.0001}.$$

Since,

$$\sqrt{0.0001} = 0.01,$$

we have:

$$T = 2\pi \times 0.01 = 0.02\pi\ \text{s}.$$

The problem states that the time period is given by:

$$y \pi \times 10^{-2}\ \text{s}.$$

Writing our result in the same form:

$$0.02\pi\ \text{s} = 2\pi \times 10^{-2}\ \text{s}.$$

We see that $$y = 2.$$

Therefore, the correct answer is:

Option A: 2.