A D-aldotetrose on oxidation with concentrated HNO$$_3$$ resulted in optically inactive dicarboxylic acid. The structure of the D-aldotetrose is:

Aldotetroses contain two chiral centres (at C-2 and C-3). Hence each D-aldotetrose has a Fischer projection of the type

$$\begin{array}{c} \mathrm{CHO}\\ \mathrm{C_2:}\;H\;|\;OH\\ \mathrm{C_3:}\;H\;|\;OH\\ \mathrm{CH_2OH} \end{array}$$

The two possible D-isomers are

$$\textbf{D-erythrose}:\;HO{-}CH{-}CH{-}OH $$ (OH groups on the same side)
$$\textbf{D-threose}:\;HO{-}CH{-}CH{-}OH $$ (OH groups on opposite sides)

Hot concentrated $$\text{HNO}_3$$ is a strong oxidising agent; it converts the aldehyde group at C-1 and the primary alcohol at C-4 into carboxylic groups, giving the corresponding aldaric acid (a dicarboxylic acid).

$$\mathrm{HOOC{-}(CHOH)_2{-}COOH}$$

After oxidation:

Case 1: Starting from D-erythrose, both chiral centres retain the original (R,S) configuration. The product is $$\textbf{meso-tartaric acid}$$, which possesses an internal plane of symmetry. Because of this internal compensation it is optically inactive. Case 2: Starting from D-threose, the configurations become (R,R). The product is $$\mathbf{d\!-\!tartaric\;acid}$$, which has no plane of symmetry and therefore is optically active.

The question states that the dicarboxylic acid obtained is optically inactive, so the parent aldotetrose must be the one that furnishes meso-tartaric acid, namely $$\textbf{D-erythrose}$$.

Among the given figures, Option (c) corresponds to the Fischer projection of D-erythrose (OH on the right at both C-2 and C-3).

Hence the required structure is furnished by
Option C which is: Option (c)