Comment to @SK Dash - quantum fluctuations are not thermal bath (how do they distribute and coubple to various degrees of freedom?), so the associated energy scale can hardly be called temperature.

The equation you wrote is more of a definition for heat capacity C ($Q = \Delta E = \frac{\partial E}{\partial T} \Delta T$), unrelated to "vacuum temperature". Massless particles, such as photons and phonons, also have finite heat capacity (for example see here and here).

Temperature is one of the least understood concepts in physics, even by physicists.

The reason the inter-galactic vacuum has finite temperature is since the EM radiation permeates it (residuals from the big-bang), and distributes as black body radiation. This EM radiation acts as the thermal bath for matter that's there. But if somehow you manage to remove this radiation (it's possible to cut partially eliminate it, making it non-thermal distribution) - the answer changes dramatically (0 for no radiation or undefined for partial distribution)

Variations of this question were asked previously here, here and here (there you can see other struggle for consistent answer). Many more on the internet.

Bottom line, the question you asked is not completely defined, so several answers exist (0, 3K, undefined and more). Search "Vacuum temperature".

Comment to @SK Dash - quantum fluctuations are not thermal bath (how do they distribute and couple to various degrees of freedom?), so the associated energy scale can hardly be called temperature.

The equation you wrote is more of a definition for heat capacity C ($Q = \Delta E = \frac{\partial E}{\partial T} \Delta T$), unrelated to "vacuum temperature". Massless particles, such as photons and phonons, also have finite heat capacity (for example see here and here).

Temperature is one of the least understood concepts in physics, even by physicists.

The reason the inter-galactic vacuum has finite temperature is since the EM radiation permeates it (residuals from the big-bang), and distributes as black body radiation. This EM radiation acts as the thermal bath for matter that's there. But if somehow you manage to remove this radiation (it's possible to cut portion of it out, making it non-thermal distribution) - the answer changes dramatically (0 for no radiation or undefined for partial distribution)

Variations of this question were asked previously here, here and here (there you can see other struggle for consistent answer). Many more on the internet.

Bottom line, the question you asked is not completely defined, so several answers exist (0, 3K, undefined and more). Search "Vacuum temperature".