X = 3YZ² find dimension of Y in (MKSA) system, if X and Z are the dimension of capacity and magnetic field respectively.

The given formula is:
$X=3Y{Z}^2$
We need to find the dimensions of $Y$ in the MKSA (Meter-Kilogram-Second-Ampere) system, where $X$ has the dimensions of capacity (electrical capacitance) and $Z$ has the dimensions of a magnetic field.

First, let us determine the dimensions of capacity, $X$:
Capacitance is defined as $C=\frac{Q}{V}$, where $Q$ is charge and $V$ is potential (voltage).
Since charge is current multiplied by time, $\left[Q\right]=AT$.
Potential $V$ is work done per unit charge, $V=\frac{W}{Q}$. The dimensions of work $W$ (energy) are $\left[W\right]=M{L}^2{T}^{-2}$.
Therefore, the dimensions of $V$ are:
$\left[V\right]=\frac{M{L}^2{T}^{-2}}{AT}=M{L}^2{T}^{-3}{A}^{-1}$
Thus, the dimensions of capacitance $X$ are:
$\left[X\right]=\frac{AT}{M{L}^2{T}^{-3}{A}^{-1}}=M^{-1}{L}^{-2}{T}^4{A}^2$

Next, let us determine the dimensions of the magnetic field, $Z$:
We use the magnetic force formula on a moving charge: $F=qvB\Rightarrow B=\frac{F}{qv}$, where $F$ is force, $q$ is charge, and $v$ is velocity.
The dimensions of force are $\left[F\right]=ML{T}^{-2}$, charge is $\left[q\right]=AT$, and velocity is $\left[v\right]=L{T}^{-1}$.
Thus, the dimensions of the magnetic field $Z$ are:
$\left[Z\right]=\frac{ML{T}^{-2}}{\left(AT\right)\left(L{T}^{-1}\right)}=M{T}^{-2}{A}^{-1}$
The square of the dimensions of $Z$ is:
$\left[Z^2\right]=M^2{T}^{-4}{A}^{-2}$

Now, we rearrange the original formula to solve for $Y$. The constant 3 is dimensionless, so:
$\left[Y\right]=\frac{\left[X\right]}{\left[Z^2\right]}$
Substituting the dimensional formulas we found:
$\left[Y\right]=\frac{M^{-1}{L}^{-2}{T}^4{A}^2}{M^2{T}^{-4}{A}^{-2}}$
Simplifying the powers for each base unit:
For M: $-1-2=-3$
For L: $-2$
For T: $4-(-4)=8$
For A: $2-(-2)=4$

Therefore, the dimensional formula for $Y$ is:
$\left[Y\right]=M^{-3}{L}^{-2}{T}^8{A}^4$