The quantities $$x = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$$, $$y = \frac{E}{B}$$ and $$z = \frac{l}{CR}$$ are defined where C-capacitance, R-Resistance, $$l$$-length, E-Electric field, B-magnetic field and $$\varepsilon_0$$, $$\mu_0$$ - free space permittivity and permeability respectively. Then:

First we recall the symbols for the fundamental (base) dimensions that we will use throughout the solution:

$$$[M] \;{\rm for\; mass},\qquad [L] \;{\rm for\; length},\qquad [T] \;{\rm for\; time},\qquad [I] \;{\rm for\; electric\; current}.$$$

We now determine the dimensions of the two electromagnetic constants that occur in the question.

For the permeability of free space we start from the defining relation $$B \;=\; \mu_0\,H.$$ Here $$B$$ is the magnetic flux density and $$H$$ is the magnetic field intensity. In SI units $$B$$ (tesla) has the dimension $$$[B] \;=\; \frac{{\rm force}}{{\rm current}\times{\rm length}} \;=\; \frac{[M\,L\,T^{-2}]}{[I]\,[L]} \;=\; [M\,T^{-2}\,I^{-1}],$$$ and $$H$$ (ampere per metre) has the dimension $$[H]\;=\;[I\,L^{-1}].$$ Therefore $$$[\mu_0] \;=\; \frac{[B]}{[H]} \;=\; \frac{[M\,T^{-2}\,I^{-1}]}{[I\,L^{-1}]} \;=\; [M\,L\,T^{-2}\,I^{-2}].$$$

For the permittivity of free space we employ Coulomb’s law $$F \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r^2},$$ so that $$\varepsilon_0 \;=\; \frac{q_1q_2}{4\pi\,F\,r^2}.$$ Ignoring numerical constants, its dimension is $$$[\varepsilon_0] \;=\; \frac{[Q]^2}{[M\,L\,T^{-2}]\,[L]^2} \;=\; \frac{[I^2\,T^2]}{[M\,L^3\,T^{-2}]} \;=\; [M^{-1}\,L^{-3}\,T^{4}\,I^{2}].$$$

With these preliminaries completed, we are ready to analyse each of the three given quantities.

1. Dimension of $$x=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}.$$

First multiply the dimensions of $$\mu_0$$ and $$\varepsilon_0$$:

$$$[\mu_0\varepsilon_0] \;=\;[M\,L\,T^{-2}\,I^{-2}]\times[M^{-1}\,L^{-3}\,T^{4}\,I^{2}] \;=\;L^{(1-3)}\,T^{(-2+4)}\,M^{(1-1)}\,I^{(-2+2)} \;=\;[L^{-2}\,T^{2}].$$$

The square root of this product therefore has the dimension

$$\sqrt{[\mu_0\varepsilon_0]} \;=\;[L^{-1}\,T^{1}].$$

Because $$x$$ is the reciprocal of that square root, we invert the above result:

$$[x]\;=\;\frac{1}{[L^{-1}\,T^{1}]}\;=\;[L^{1}\,T^{-1}].$$

This is exactly the dimension of speed (length per time).

2. Dimension of $$y=\dfrac{E}{B}.$$

We already know $$[B]=[M\,T^{-2}\,I^{-1}].$$ Let us now find $$[E]$$, the electric field intensity. By definition $$E \;=\; \frac{\text{force}}{\text{charge}},$$ so $$$[E] \;=\;\frac{[M\,L\,T^{-2}]}{[I\,T]} \;=\;[M\,L\,T^{-3}\,I^{-1}].$$$

Dividing $$E$$ by $$B$$ gives

$$$[y] \;=\;\frac{[M\,L\,T^{-3}\,I^{-1}]}{[M\,T^{-2}\,I^{-1}]} \;=\;[L\,T^{-1}].$$$

Thus $$y$$ also has the dimension of speed.

3. Dimension of $$z=\dfrac{l}{C\,R}.$$

The length $$l$$ has the obvious dimension $$[l]=[L].$$

For the capacitance $$C$$ we use the definition $$C \;=\; \frac{Q}{V},$$ and voltage (potential difference) satisfies $$$V \;=\; \frac{\text{work}}{\text{charge}} \;=\; \frac{[M\,L^{2}\,T^{-2}]}{[I\,T]} \;=\;[M\,L^{2}\,T^{-3}\,I^{-1}].$$$ Hence

$$$[C] \;=\;\frac{[I\,T]}{[M\,L^{2}\,T^{-3}\,I^{-1}]} \;=\;[M^{-1}\,L^{-2}\,T^{4}\,I^{2}].$$$

For resistance $$R$$, Ohm’s law gives $$R = V/I,$$ so

$$$[R] \;=\;\frac{[M\,L^{2}\,T^{-3}\,I^{-1}]}{[I]} \;=\;[M\,L^{2}\,T^{-3}\,I^{-2}].$$$

The product $$C\,R$$ therefore possesses the dimension

$$$[C\,R] \;=\;[M^{-1}\,L^{-2}\,T^{4}\,I^{2}] \times[M\,L^{2}\,T^{-3}\,I^{-2}] \;=\;L^{(-2+2)}\,T^{(4-3)}\,M^{(-1+1)}\,I^{(2-2)} \;=\;[T^{1}].$$$

Finally, dividing the length $$l$$ by this product yields

$$[z] \;=\;\frac{[L]}{[T]} \;=\;[L\,T^{-1}].$$

Thus $$z$$ again has the dimension of speed.

We have shown that

$$$[x]=[L\,T^{-1}],\qquad [y]=[L\,T^{-1}],\qquad [z]=[L\,T^{-1}].$$$

All three quantities share the identical dimension (that of velocity).

Hence, the correct answer is Option A.