Match List I with List II.
    List-I                                                             List-II
a. Capacitance, C                                  i.   $$M^1 L^1 T^{-3} A^{-1}$$
b. Permittivity of free space, $$\varepsilon_0$$          ii.  $$M^{-1} L^{-3} T^4 A^2$$
c. Permeability of free space, $$\mu_0$$       iii. $$M^{-1} L^{-2} T^4 A^2$$
d. Electric field, E                                  iv. $$M^1 L^1 T^{-2} A^{-2}$$
Choose the correct answer from the options given below:

We have to find the dimensional formulas of the four physical quantities of List I and then compare them with the four possibilities given in List II. The dimensional symbols used are $$M$$ for mass, $$L$$ for length, $$T$$ for time and $$A$$ for electric current. After obtaining each formula we will match it with the corresponding item (i, ii, iii, iv) given in the problem.

Capacitance $$C$$
The defining relation is $$C=\dfrac{Q}{V}$$ where $$Q$$ is charge and $$V$$ is potential difference.
First, the dimension of charge is written from $$Q = I\,t,$$ so  $$[Q]=A\,T.$$ Next, the potential difference $$V$$ is work done per unit charge, $$V=\dfrac{W}{Q}.$$ Work (or energy) has the mechanical dimensions of force into distance, so $$[W]=M\,L^{2}\,T^{-2}.$$ Dividing by charge gives
$$[V]=\dfrac{M\,L^{2}\,T^{-2}}{A\,T}=M\,L^{2}\,T^{-3}\,A^{-1}.$$ Now substitute these in the formula for $$C$$:
$$[C]=\dfrac{A\,T}{M\,L^{2}\,T^{-3}\,A^{-1}} =A\,T\;M^{-1}\,L^{-2}\,T^{3}\,A^{1} =M^{-1}\,L^{-2}\,T^{4}\,A^{2}.$$ Thus capacitance corresponds to $$M^{-1}L^{-2}T^{4}A^{2},$$ which is item (iii) of List II.

Permittivity of free space $$\varepsilon_{0}$$
Coulomb’s law states $$F=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q_{1}q_{2}}{r^{2}}.$$ Rearranging gives
$$\varepsilon_{0}=\dfrac{q_{1}q_{2}}{4\pi\,r^{2}\,F}.$$
Ignoring the numerical constant $$4\pi,$$ the dimensional expression is
$$[\varepsilon_{0}]=\dfrac{(A\,T)^{2}}{L^{2}\,(M\,L\,T^{-2})} =\dfrac{A^{2}\,T^{2}}{M\,L^{3}\,T^{-2}} =M^{-1}\,L^{-3}\,T^{4}\,A^{2}.$$ Hence $$\varepsilon_{0}$$ matches item (ii).

Permeability of free space $$\mu_{0}$$
For two long, straight, parallel conductors the magnetic force per unit length is
$$\dfrac{F}{l}=\dfrac{\mu_{0}}{2\pi}\dfrac{I_{1}I_{2}}{r}.$$
Rearranging yields $$\mu_{0}=\dfrac{F/l\;\,2\pi\,r}{I^{2}}.$$ Omitting the constant $$2\pi$$ we have
$$[\mu_{0}]=\dfrac{M\,T^{-2}\,L}{A^{2}} =M\,L\,T^{-2}\,A^{-2}.$$ Therefore $$\mu_{0}$$ corresponds to item (iv).

Electric field $$E$$
The electric field is force per charge: $$E=\dfrac{F}{Q}.$$ Using $$[F]=M\,L\,T^{-2}$$ and $$[Q]=A\,T,$$ we get
$$[E]=\dfrac{M\,L\,T^{-2}}{A\,T}=M\,L\,T^{-3}\,A^{-1}.$$ This equals item (i).

Collecting the results:
a. $$C$$ → (iii)    b. $$\varepsilon_{0}$$ → (ii)    c. $$\mu_{0}$$ → (iv)    d. $$E$$ → (i)

This sequence (a-iii, b-ii, c-iv, d-i) is exactly the pairing given in Option A.

Hence, the correct answer is Option A.