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

We have the dimensional equation $$X = 5\,Y\,Z^{2}$$. The numeral 5 is dimensionless, so the dimensions of the physical quantities must satisfy

$$[X] = [Y]\,[Z]^{2}\,.$$

Hence, to find the dimensions of $$Y$$ we need $$[Y] = \dfrac{[X]}{[Z]^{2}}.$$

Step 1 - Dimensions of $$X$$ (Capacitance)
Capacitance $$C$$ is defined by the relation $$C = \dfrac{Q}{V}$$, where $$Q$$ is electric charge and $$V$$ is potential difference.

The dimension of charge is current × time: $$[Q] = A\,T.$$

Voltage is work per unit charge: $$V = \dfrac{W}{Q}.$$ Work (or energy) has the dimensions of $$M\,L^{2}\,T^{-2}$$, so

$$[V] = \dfrac{M\,L^{2}\,T^{-2}}{A\,T} = M\,L^{2}\,T^{-3}\,A^{-1}\,.$$

Therefore, the dimensional formula of capacitance is

$$[C] = \dfrac{A\,T}{M\,L^{2}\,T^{-3}\,A^{-1}} = M^{-1}\,L^{-2}\,T^{4}\,A^{2}\,.$$

Since $$X$$ has the dimensions of capacitance, we write

$$[X] = M^{-1}\,L^{-2}\,T^{4}\,A^{2}\,.$$

Step 2 - Dimensions of $$Z$$ (Magnetic field)
Magnetic field $$B$$ is defined through the Lorentz force $$F = q\,v\,B$$ on a moving charge. Rearranging, $$B = \dfrac{F}{q\,v}.$$

The force dimension is $$[F] = M\,L\,T^{-2}\,,$$ charge is $$[q] = A\,T\,,$$ and speed is $$[v] = L\,T^{-1}\,.$$ Therefore

$$[B] = \dfrac{M\,L\,T^{-2}}{A\,T \, \cdot \, L\,T^{-1}} = \dfrac{M\,L\,T^{-2}}{A\,L} = M\,T^{-2}\,A^{-1}\,.$$

This is the dimensional formula of a magnetic field (tesla). Thus

$$[Z] = M\,T^{-2}\,A^{-1}\,.$$

Step 3 - Square of $$Z$$ and its inverse

Squaring $$Z$$ gives

$$[Z]^{2} = (M\,T^{-2}\,A^{-1})^{2} = M^{2}\,T^{-4}\,A^{-2}\,.$$

The reciprocal of this quantity, which we need for $$\dfrac{1}{[Z]^{2}}$$, is

$$[Z]^{-2} = M^{-2}\,T^{4}\,A^{2}\,.$$

Step 4 - Dimensions of $$Y$$

Now substitute the results into $$[Y] = \dfrac{[X]}{[Z]^{2}}$$:

$$[Y] = (M^{-1}\,L^{-2}\,T^{4}\,A^{2}) \times (M^{-2}\,T^{4}\,A^{2}).$$

Combine the powers of each fundamental unit:

Mass: $$(-1) + (-2) = -3,$$

Length: $$(-2) + 0 = -2,$$

Time: $$4 + 4 = 8,$$

Current: $$2 + 2 = 4.$$

Thus

$$[Y] = M^{-3}\,L^{-2}\,T^{8}\,A^{4}\,.$$

Comparing with the given options, this matches Option D.

Hence, the correct answer is Option D.