The reaction of NO with $$\text{N}_2\text{O}_4$$ at 250 K gives:

We are given that nitric oxide is reacting with dinitrogen tetroxide at a low temperature of about 250 K. At this temperature dinitrogen tetroxide exists mainly as the dimer of nitrogen dioxide, but it is in equilibrium with a small amount of $$\text{NO}_2$$ according to

$$\text{N}_2\text{O}_4 \;\rightleftharpoons\; 2\,\text{NO}_2 \qquad (1)$$

Next we recall the well-known combination reaction between nitric oxide and nitrogen dioxide. First we state the equation and then use it:

Formula used: “One mole of $$\text{NO}$$ combines with one mole of $$\text{NO}_2$$ to give one mole of dinitrogen trioxide $$\text{N}_2\text{O}_3$$.”

$$\text{NO} \;+\; \text{NO}_2 \;\longrightarrow\; \text{N}_2\text{O}_3 \qquad (2)$$

Now we supply nitric oxide to the equilibrium mixture (1). The moment $$\text{NO}$$ appears, it finds $$\text{NO}_2$$ (generated from the slight dissociation of $$\text{N}_2\text{O}_4$$) and reacts via equation (2) to give $$\text{N}_2\text{O}_3$$. Because $$\text{NO}_2$$ is being removed, Le Châtelier’s principle makes reaction (1) shift to the right, furnishing still more $$\text{NO}_2$$, which in turn keeps combining with more $$\text{NO}$$. Combining equations (1) and (2) algebraically gives us the overall stoichiometry.

First write twice equation (2) so that $$2\,\text{NO}_2$$ are used:

$$2\,\text{NO} + 2\,\text{NO}_2 \;\longrightarrow\; 2\,\text{N}_2\text{O}_3 \qquad (2')$$

Now add equation (1) and equation (2') term by term.

Left-hand side: $$\text{N}_2\text{O}_4 + 2\,\text{NO}$$

Right-hand side: $$2\,\text{NO}_2 + 2\,\text{N}_2\text{O}_3$$

Because $$2\,\text{NO}_2$$ appears on both sides, it cancels out, leaving the net equation

$$\boxed{\,2\,\text{NO} \;+\; \text{N}_2\text{O}_4 \;\longrightarrow\; 2\,\text{N}_2\text{O}_3\,} \qquad (3)$$

This overall reaction tells us unambiguously that the species obtained is dinitrogen trioxide, $$\text{N}_2\text{O}_3$$.

Hence, the correct answer is Option C.