Step 1 - Note the conditions of nitration.
Nitration is carried out with a nitrating mixture, conc. $HNO_3$ + conc. $H_2SO_4$,
which is STRONGLY ACIDIC.

Step 2 - See what happens to aniline in that medium.
Aniline is a base. In the strongly acidic medium it is largely PROTONATED to the
ANILINIUM ION:
$$C_6H_5NH_2 + H^+ \rightarrow C_6H_5\overset{+}{N}H_3$$

Step 3 - Compare the directing effects.
$-NH_2$ (in free aniline) : activating, ORTHO/PARA directing.
$-\overset{+}{N}H_3$ (in anilinium ion) : strongly deactivating because of its positive
charge, and META directing.
Both species are present together in the reaction mixture, so substitution occurs at
ortho, para AND meta positions.

Step 4 - Give the actual composition.
The observed product distribution is approximately
$$51\%\ \text{meta}, \quad 47\%\ \text{ortho}, \quad 2\%\ \text{para}$$
The unusually high meta percentage is the tell-tale sign that the reaction proceeds
largely through the anilinium ion.

Answer: (B) 51%-meta, 47%-ortho, 2%-para derivatives