Match List - I with List - II:
        List - I                                    List - II
a. Magnetic induction        i.   $$ML^2T^{-2}A^{-1}$$
b. Magnetic flux                  ii.  $$M^0L^{-1}A$$
c. Magnetic permeability   iii. $$MT^{-2}A^{-1}$$
d. Magnetization                 iv.  $$MLT^{-2}A^{-2}$$
Choose the most appropriate answer from the options given below:

We have to match each physical quantity in List - I with its dimensional formula in List - II. Let us determine the dimensions one by one, showing every algebraic step.

(a) Magnetic induction $$\mathbf B$$
The force on a moving charge is $$\mathbf F = q\,\mathbf v \times \mathbf B.$$
Using the basic dimensions $$$[\mathbf F] = ML T^{-2},\; [q] = AT,\; [\mathbf v] = LT^{-1},$$$ we write

$$$[\mathbf B] \;=\; \frac{[\mathbf F]}{[q]\,[\mathbf v]}\;=\; \frac{MLT^{-2}}{(AT)\,(LT^{-1})}.$$$

Simplifying the powers of $$L$$ and $$T$$:

$$[\mathbf B] = M T^{-2} A^{-1}.$$

This matches List-II item $$\text{(iii)}\; M T^{-2} A^{-1}.$$

(b) Magnetic flux $$\Phi_B$$
By definition $$\Phi_B = \mathbf B \cdot \text{Area}.$$ Hence

$$$[\Phi_B] = [\mathbf B]\,[\text{Area}] = \left(M T^{-2} A^{-1}\right)\left(L^{2}\right).$$$

Multiplying, we obtain

$$[\Phi_B] = M L^{2} T^{-2} A^{-1}.$$

This corresponds to List-II item $$\text{(i)}\; M L^{2} T^{-2} A^{-1}.$$

(c) Magnetic permeability $$\mu$$
The relation between magnetic induction and magnetic field strength is $$\mathbf B = \mu\,\mathbf H.$$ Rearranging gives $$\mu = \mathbf B / \mathbf H.$$

We already have $$[\mathbf B] = M T^{-2} A^{-1}.$$ Field strength $$\mathbf H$$ is current per unit length, so $$[\mathbf H] = A L^{-1}.$$

Therefore

$$$[\mu] = \frac{M T^{-2} A^{-1}}{A L^{-1}} = M L T^{-2} A^{-2}.$$$

This is exactly List-II item $$\text{(iv)}\; M L T^{-2} A^{-2}.$$

(d) Magnetization $$\mathbf M$$
Magnetization is magnetic moment per unit volume. Magnetic moment is current times area, so $$[\text{moment}] = A L^{2}.$$ Dividing by volume $$L^{3}$$ gives

$$[\mathbf M] = \frac{A L^{2}}{L^{3}} = A L^{-1}.$$

Since no mass term appears, we may write

$$[\mathbf M] = M^{0} L^{-1} A^{1}.$$

This equals List-II item $$\text{(ii)}\; M^{0} L^{-1} A.$$

Collecting the matches:

$$$ \begin{aligned} \text{a}\;&\to\;\text{(iii)},\quad \text{b}\;&\to\;\text{(i)},\quad \text{c}\;&\to\;\text{(iv)},\quad \text{d}\;&\to\;\text{(ii)}. \end{aligned} $$$

Option B lists exactly this correspondence. Hence, the correct answer is Option B.