**Understanding the Problem:**
We are given a sphere moving in water with a velocity of 1.6 m/s. We need to find the velocity of air in a wind tunnel that will give dynamically similar conditions. The air in the wind tunnel has a density 750 times less than water and is 60 times less viscous than water. We need to determine the velocity of the air in order to achieve dynamic similarity.

**Understanding Dynamic Similarity:**
Dynamic similarity is a concept used in fluid mechanics to compare the behavior of fluid flow in different systems. It states that if two systems have the same ratio of forces acting on them, they will have similar fluid flow behavior. In this case, we are comparing the flow of water around the first sphere to the flow of air around the second sphere in the wind tunnel.

**Using Dimensional Analysis:**
We can use dimensional analysis to determine the relationship between the forces acting on the two spheres. The forces involved in this problem are the drag force and the gravitational force.

The drag force can be expressed as:
Drag force = 0.5 * rho * A * Cd * V^2

Where:
- rho is the density of the fluid
- A is the cross-sectional area of the sphere
- Cd is the drag coefficient
- V is the velocity of the fluid

The gravitational force can be expressed as:
Gravitational force = m * g

Where:
- m is the mass of the sphere
- g is the acceleration due to gravity

**Applying the Principle of Dynamic Similarity:**
To achieve dynamic similarity, the ratios of the forces acting on the two spheres should be equal.

1. Let's compare the drag forces:
Drag force1 / Drag force2 = (0.5 * rho1 * A1 * Cd1 * V1^2) / (0.5 * rho2 * A2 * Cd2 * V2^2)

Since the drag force on the second sphere is not mentioned, we assume that the drag coefficients and cross-sectional areas of both spheres are the same. Therefore, we can simplify the equation to:

rho1 / rho2 = V1^2 / V2^2

2. Let's compare the gravitational forces:
Gravitational force1 / Gravitational force2 = (m1 * g) / (m2 * g)

Since the mass and acceleration due to gravity are the same for both spheres, this ratio simplifies to:

1 = 1

**Applying the Given Conditions:**
Given that the density of air is 750 times less than water, we can substitute rho1 / rho2 with 1/750. Solving for V2, we get:

1/750 = (1.6^2) / (V2^2)

V2^2 = (1.6^2) / (1/750) = 1.6^2 * 750 = 1920

V2 = sqrt(1920) = 43.8 m/s

**Determining the Correct Answer:**
The closest option to the calculated velocity V2 is 10 m/s, which is option B. Therefore, the velocity of air in the wind tunnel that will give dynamically similar conditions is 10 m/s.