Step 1: Concept, use the Stefan-Boltzmann law instead of a bare ratio.
The energy radiated per second per unit area by a real surface is $E = \epsilon \sigma T^4$, where $\epsilon$ is the emissivity, $\sigma$ is the Stefan-Boltzmann constant and $T$ is the absolute temperature. Rearranging gives $\epsilon = \dfrac{E}{\sigma T^4}$, so once we know the dimensional formula of $E$ and of $\sigma$, the dimension of $\epsilon$ falls out directly.
Step 2: Dimensional formula of $E$ (power per unit area).
Power has dimension $[M L^2 T^{-3}]$ (energy per time), so power per unit area is
\[ [E] = \frac{[M L^2 T^{-3}]}{[L^2]} = [M L^0 T^{-3}] \]
Step 3: Dimensional formula of the Stefan-Boltzmann constant.
From the same law written for a perfect black body, $E_b = \sigma T^4$, so $[\sigma] = \dfrac{[E_b]}{[T^4]}$. Using $[E_b] = [M T^{-3}]$ and temperature dimension $[\Theta]$ for $T$:
\[ [\sigma] = [M T^{-3} \Theta^{-4}] \]
This is a standard constant worth remembering on its own, not just something you back out of a ratio.
Step 4: Combine to get the dimension of emissivity.
\[ [\epsilon] = \frac{[M T^{-3}]}{[M T^{-3} \Theta^{-4}][\Theta^4]} = \frac{[M T^{-3}]}{[M T^{-3}]} = [M^0 L^0 T^0] \]
The mass, time and temperature powers all cancel exactly, confirming emissivity carries no dimension at all.
Step 5: Rule out the other options.
$[M^1 L^0 T^{-3}]$ and $[M^1 L^2 T^{-3}]$ still carry a mass dimension, so they describe power or power per area, not a dimensionless ratio; $[M^1 L^2 T^{-2}]$ is the dimension of energy itself. None of these can be right for a quantity defined purely as a ratio of two powers of the same kind.
Final Answer:
Emissivity is dimensionless because it is the ratio of two quantities of the same physical nature.
\[ \boxed{[M^0 L^0 T^0]} \]
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