In the formula  $X=5Y{Z}^2$, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y?

$$\begin{gathered} X=5Y{Z}^2 \\ \text{X=capacitance=}\frac{Q}{V} \\ =\frac{Q^2}{\omega }=\frac{Q^2}{F.L}=\frac{A^2{T}^2}{M^1{L}^1{T}^{-2}\times L}=A^2{M}^{-1}{L}^{-2}{T}^4 \\ Z=B=\frac{F}{i\mathcal{l}}=\frac{M^1{L}^1{T}^{-2}}{A^1{L}^1}=M^1{L}^0{T}^{-2}{A}^{-1} \\ {Z}^2=M^2{T}^{-4}{A}^{-2} \\ Y=\frac{X}{Z^2}=\frac{M^{-1}{L}^{-2}{T}^4{A}^2}{M^2{T}^{-4}{A}^{-2}} \\ =M^{-3}{L}^{-2}{T}^8{A}^4 \end{gathered}$$