1. Identify the reaction of $Cr^{3+}$ with dilute $NaOH$

When a salt solution containing $Cr^{3+}$ is treated with dilute $NaOH$, hydroxide ions are added to the aqueous chromium(III) ions.

Initially: $Cr^{3+} + 3OH^- \rightarrow Cr(OH)_3 \downarrow$

So a gelatinous precipitate is formed.

2. Nature of the precipitate

In qualitative inorganic chemistry, the precipitated $Cr(OH)_3$ is often represented more realistically as a hydrated oxide: $Cr_2O_3\cdot nH_2O$

Thus, the precipitate obtained from adding dilute $NaOH$ to $Cr^{3+}$ solution is commonly written as: $Cr_2O_3(H_2O)_n$

3. Check the options

• A: a solution of $[Cr(OH)_4]^-$
  This is not obtained on adding only dilute $NaOH$. Chromium(III) hydroxide is amphoteric, but dissolution requires excess alkali.

• B: precipitate of $Cr_2O_3(H_2O)_n$
  Correct. This is the hydrated form of precipitated chromium(III) hydroxide.

• C: precipitate of $[Cr(OH)_6]^{3-}$
  Incorrect. This is a complex ion, not a precipitate.

• D: precipitate of $Cr(OH)_3$
  Chemically this also describes the same precipitate, but in standard JEE-type treatment the precipitate is more specifically represented as hydrated chromium(III) oxide, i.e. $Cr_2O_3(H_2O)_n$.

4. Final choice

Hence, the best answer is: $\boxed{B}$

5. Comparison with stored correct answer

Stored correct answer = B
Derived answer = B
So they agree.