To solve this problem, we need to find the dimensions of the electric flux and the magnetic flux, and then take their ratio.

Determine the dimension of electric flux (Φₑ):

Electric flux is defined as

$$\Phi_E = \int_S \mathbf{E} \cdot d\mathbf{A}.$$

The electric field $\mathbf{E}$ has dimensions:

$$[E] = \frac{\text{Force}}{\text{Charge}}.$$

Since force has dimensions $$M L T^{-2},$$ and charge (in SI) has dimensions $$A\,T,$$ we get:

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

The area element $$dA$$ has dimensions $$L^2.$$

Thus, the electric flux has dimensions:

$$[\Phi_E] = [E][A] = M L T^{-3} A^{-1} \times L^2 = M L^3 T^{-3} A^{-1}.$$

Determine the dimension of magnetic flux (Φ_B):

Magnetic flux is defined as

$$\Phi_B = \int_S \mathbf{B} \cdot d\mathbf{A}.$$

The magnetic field $\mathbf{B}$ has dimensions determined via the Lorentz force law (using $$F = q\,v\,B$$), giving:

$$[B] = \frac{F}{q\,v} = \frac{M L T^{-2}}{(A\,T)(L T^{-1})} = M T^{-2} A^{-1}.$$

Again, the area element has dimensions $$L^2.$$

So the magnetic flux has dimensions:

$$[\Phi_B] = [B][A] = (M T^{-2} A^{-1})(L^2) = M L^2 T^{-2} A^{-1}.$$

Find the ratio of electric flux to magnetic flux:

Taking the ratio:

$$\frac{\Phi_E}{\Phi_B} = \frac{M L^3 T^{-3} A^{-1}}{M L^2 T^{-2} A^{-1}}.$$

Canceling like terms:

Mass $(M)$ cancels.

Current $(A^{-1})$ cancels.

For length, $$L^{3-2} = L^1.$$

For time, $$T^{-3 -(-2)} = T^{-1}.$$

Therefore, the ratio has dimensions:

$$L^{1}T^{-1}.$$

This means in the expression

$$M^P L^Q T^R A^S,$$

the exponents for length and time are $$Q = 1$$ and $$R = -1.$$

Looking at the options provided, this corresponds to Option D: $$(1,-1).$$

Thus, the answer is Option D.