1. Identify the transformation

We need to convert toluene 4benzyl methyl group4 into benzaldehyde:

$C_6H_5CH_3 \longrightarrow C_6H_5CHO$

This is controlled side-chain oxidation of an alkylbenzene.

2. Check each reagent

Option A: $CrO_3$/acetic acid, $H_3O^+$

This is a strong oxidizing system. Such reagents generally oxidize the benzylic side chain further, not stopping cleanly at the aldehyde stage. The product tends toward benzoic acid rather than benzaldehyde.

So, A is not suitable.

Option B: $CrO_3$/acetic anhydride, $H_3O^+$

This is the Etard-type controlled oxidation condition used for converting toluene to benzaldehyde. In acetic anhydride, overoxidation is suppressed, and after hydrolysis:

$C_6H_5CH_3 \xrightarrow[H_3O^+]{CrO_3/(CH_3CO)_2O} C_6H_5CHO$

So, B is correct.

Option C: $KMnO_4$/$HCl$, $H_3O^+$

$KMnO_4$ is a very strong oxidizing agent for benzylic side chains. Toluene is oxidized completely to benzoic acid:

$C_6H_5CH_3 \xrightarrow{KMnO_4} C_6H_5COOH$

Hence, C is incorrect.

Option D: $CO$/$HCl$, anhydrous $AlCl_3$

This is the GattermannKoch formylation reagent. It introduces a formyl group directly onto benzene rings by electrophilic substitution. It is used for:

$C_6H_6 \rightarrow C_6H_5CHO$

It is not oxidation of toluene to benzaldehyde.

So, D is incorrect.

3. Conclusion

The reagent that easily oxidizes toluene to benzaldehyde is:

$\boxed{\text{B: } CrO_3/\text{acetic anhydride},\ H_3O^+}$