1. Given data

• Cross-sectional area: $A = 2\,\text{cm}^2 = 2 \times 10^{-4}\,\text{m}^2$
• Length of rod: $l = 40\,\text{cm} = 0.4\,\text{m}$
• Number of turns: $N = 400$
• Current: $I = 0.4\,\text{A}$
• Magnetic flux inside windings: $\phi = 4\pi \times 10^{-6}\,\text{Wb}$
• Permeability of free space: $\mu_0 = 4\pi \times 10^{-7}\,\text{N A}^{-2}$

We need to find the relative permeability $\mu_r$ of the rod.

2. Magnetic field inside a solenoid with core

For a uniformly wound rod (solenoid-like arrangement),

$B = \mu_0 \mu_r \frac{NI}{l}$

Also,

$\phi = BA$

So,

$B = \frac{\phi}{A}$

Hence,

$\frac{\phi}{A} = \mu_0 \mu_r \frac{NI}{l}$

Therefore,

$\mu_r = \frac{\phi l}{A\mu_0 NI}$

3. Substitute the values

$\mu_r = \frac{(4\pi \times 10^{-6})(0.4)}{(2\times 10^{-4})(4\pi \times 10^{-7})(400)(0.4)}$

We can cancel $0.4$ from numerator and denominator:

$\mu_r = \frac{4\pi \times 10^{-6}}{(2\times 10^{-4})(4\pi \times 10^{-7})(400)}$

Now simplify the denominator:

$(2\times 10^{-4})(400) = 8\times 10^{-2}$

So,

$\text{denominator} = (8\times 10^{-2})(4\pi \times 10^{-7}) = 32\pi \times 10^{-9}$

Thus,

$\mu_r = \frac{4\pi \times 10^{-6}}{32\pi \times 10^{-9}}$

$\mu_r = \frac{4}{32} \times 10^3 = \frac{1}{8}\times 1000 = 125$

4. Check options

• A: $\frac{5}{16}$
• B: $125$
• C: $\frac{32}{5}$
• D: $12.5$

So the correct option is:

$\boxed{\text{B: }125}$

5. Comparison with stored answer

Stored correct answer is A, but the calculated value is clearly:

$\boxed{125}$

Hence the stored answer appears to be incorrect.