The question says:

A body $X$ of emissivity $\epsilon$ is at temperature $T_1$ . $X$ is inside a box whose walls act as a black body of temperature $T_2$ . $T_1$ is greater than $T_2$ . What is the net intensity of radiation leaving body $X?$

The answer states that it is $I_{\text{net}}=\epsilon \sigma (T_1^4-T_2^4)$, but my question is why isn't it $I_{\text{net}}=\epsilon \sigma (T_1^4+T_2^4)$, because the body $X$ absorbs the radiation from the walls of temperature $T_2,$ so it needs to re radiate this power, and therefore should be an increase in the net intensity. What am I missing here with my understanding?

Don't confuse the instantaneous emission of the object (which depends only on its temperature) with the net flow (which is the difference between the inflow and outflow).

Yes an inflow could potentially add to the object's energy. But it would do so by changing the temperature. So we don't have to worry about that for the instantaneous flow.

At an instant of time you are evaluating the net rate of flow of energy between the two bodies.

What happens next depends on the characteristics of the two bodies.
It might be that because of the net flow of energy the body which had a temperature $T_1$ now has a lower temperature $T’_1$ and the body which had a temperature $T_2$ now has a higher temperature $T’_2$ and then these two new temperatures will dictate the instantaneous net rate of flow of energy between the two bodies.

If the temperature of the two bodies stay the same then the instantaneous net rate of flow of energy will stay the same.

Your suggestion of an increase in the rate at which energy is radiated by the hot body implies that it is now at a higher temperature and correspondingly the other body is at a lower temperature, but the second law of thermodynamics does not allow heat to flow from a colder body to a hotter body.

The question asks about the net radiation of X, not of the whole system/scene. If blackbody X radiates $\psi_1$, and it absorbs radiation $\psi_2$, the net radiation of X is $\psi_{X}=\psi_1-\psi_2$. For the sake of this question you can assume that the effect $T_2$ would have on $T_1$ is already included in $T_1$ (the implied increase in temperature).