1. Target compound analysis

We need to prepare 3-nitrobenzoic acid from benzene.

The target has:

• a carboxylic acid group $(-COOH)$
• a nitro group $(-NO_2)$
• and the nitro group is meta to the carboxylic acid.

So we must choose a sequence that gives the meta relationship correctly.

2. Directing effects to keep in mind

• $-NO_2$ is a strong deactivating, meta-directing group.
• $-COOH$ is also meta-directing, but introducing $-COOH$ directly from benzene is not straightforward.
• Halogens like $Br$ are deactivating but ortho/para-directing.

Thus, to obtain $m$-nitrobenzoic acid, a good idea is:

• first introduce $NO_2$ on benzene,
• then introduce $Br$ meta to $NO_2$,
• convert that $Br$ into $COOH$ using Grignard carboxylation.

3. Check each option

Option A

$Br_2/AlBr_3 \rightarrow HNO_3/H_2SO_4 \rightarrow Mg/ether \rightarrow CO_2 \rightarrow H_3O^+$

Stepwise:

• Benzene $\xrightarrow{Br_2/AlBr_3}$ bromobenzene
• Bromobenzene $\xrightarrow{HNO_3/H_2SO_4}$ nitration

Since $Br$ is ortho/para-directing, nitration gives mainly o- and p-nitrobromobenzene, not meta.

Then conversion of $Br$ to Grignard followed by $CO_2/H_3O^+$ would replace $Br$ by $COOH$. So products become o- or p-nitrobenzoic acid, not m-nitrobenzoic acid.

Hence, A is incorrect.

Option B

$Br_2/AlBr_3 \rightarrow NaCN \rightarrow H_3O^+ \rightarrow HNO_3/H_2SO_4$

After bromination, we get bromobenzene. But aryl halides do not undergo simple nucleophilic substitution with $NaCN$ under ordinary conditions. So bromobenzene cannot be directly converted to benzonitrile this way.

Therefore, B is incorrect.

Option C

$Br_2/AlBr_3 \rightarrow HNO_3/H_2SO_4 \rightarrow NaCN \rightarrow H_3O^+$

Again:

• bromination gives bromobenzene
• nitration gives mainly ortho/para nitrobromobenzene
• then $NaCN$ cannot simply replace aryl $Br$

So this is invalid both in orientation and reactivity.

Hence, C is incorrect.

Option D

$HNO_3/H_2SO_4 \rightarrow Br_2/AlBr_3 \rightarrow Mg/ether \rightarrow CO_2 \rightarrow H_3O^+$

Stepwise:

Step 1: Nitration

$C_6H_6 \xrightarrow{HNO_3/H_2SO_4} C_6H_5NO_2$ Benzene gives nitrobenzene.

Step 2: Bromination

$C_6H_5NO_2 \xrightarrow{Br_2/AlBr_3} m\text{-bromonitrobenzene}$ Because $NO_2$ is meta-directing, bromine enters the meta position.

Step 3: Grignard formation

$m\text{-bromonitrobenzene} \xrightarrow{Mg/ether} m\text{-nitrophenyl magnesium bromide}$

Step 4: Carboxylation

$ArMgBr \xrightarrow{CO_2} ArCOO^-MgBr^+$

Step 5: Acidic hydrolysis

$ArCOO^-MgBr^+ \xrightarrow{H_3O^+} ArCOOH$

Thus the bromine-bearing carbon is converted into $COOH$, giving: $m\text{-nitrobenzoic acid}$

So D gives the required product.

4. Final answer

The correct sequence is: $\boxed{D}$