Solution:

To solve this problem, we use the formula for the magnetic field inside a long solenoid:

$$B = \mu_0 n i,$$

where:

• $B$ is the magnetic field at the center of the solenoid,
• $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space,
• $n$ is the number of turns per unit length (in meters),
• $i$ is the current in the solenoid.

Step 1: Magnetic Field for the First Solenoid

The first solenoid has:

• $n_1 = 200 \, \text{turns per cm} = 200 \times 100 = 20000 \, \text{turns per meter}$,
• $i_1 = i$,
• $B_1 = 6.28 \times 10^{-2} \, \text{weber/m}^2$.

Using the formula for $B$, substitute $B_1$ to find $i$:

$$B_1 = \mu_0 n_1 i,$$ $$6.28 \times 10^{-2} = (4\pi \times 10^{-7}) \cdot 20000 \cdot i.$$

Simplify:

$$i = \frac{6.28 \times 10^{-2}}{4\pi \times 10^{-7} \cdot 20000}.$$

Substitute $4\pi \approx 12.57$:

$$i = \frac{6.28 \times 10^{-2}}{12.57 \times 10^{-7} \cdot 20000} = \frac{6.28 \times 10^{-2}}{2.514 \times 10^{-2}}.$$ $$i = 2.5 \, \text{A}.$$

Step 2: Magnetic Field for the Second Solenoid

The second solenoid has:

• $n_2 = 100 \, \text{turns per cm} = 100 \times 100 = 10000 \, \text{turns per meter}$,
• $i_2 = \frac{i}{3} = \frac{2.5}{3} = 0.833 \, \text{A}$.

Using the formula for $B$:

$$B_2 = \mu_0 n_2 i_2,$$

Substitute the values:

$$B_2 = (4\pi \times 10^{-7}) \cdot 10000 \cdot 0.833.$$

Simplify:

$$B_2 = 4\pi \times 10^{-7} \cdot 8330.$$

Substitute $4\pi \approx 12.57$:

$$B_2 = 12.57 \times 10^{-7} \cdot 8330 = 1.048 \times 10^{-2}.$$

Final Answer:

$$\boxed{B_2 = 1.05 \times 10^{-2} \, \text{weber/m}^2}$$