Term Independent of x in Binomial Expansion

To find the term independent of $x$ in the binomial expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$, we use the general term formula for a binomial expansion $(a + b)^n$:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Here, $n = 18$, $a = 9x$, and $b = -\frac{1}{3\sqrt{x}} = -\frac{1}{3}x^{-1/2}$.

Substituting these values into the general term formula gives:

$T_{r+1} = \binom{18}{r} (9x)^{18-r} \left(-\frac{1}{3\sqrt{x}}\right)^r$

Rewriting the constant and variable parts separately:

$T_{r+1} = \binom{18}{r} (3^2)^{18-r} \cdot x^{18-r} \cdot (-1)^r \cdot 3^{-r} \cdot x^{-r/2}$

$T_{r+1} = (-1)^r \binom{18}{r} 3^{36 - 2r - r} \cdot x^{18 - r - \frac{r}{2}}$

$T_{r+1} = (-1)^r \binom{18}{r} 3^{36 - 3r} \cdot x^{18 - \frac{3r}{2}}$

For the term to be independent of $x$, the exponent of $x$ must be zero:

$18 - \frac{3r}{2} = 0$

$\frac{3r}{2} = 18 \implies 3r = 36 \implies r = 12$

Now, substituting $r = 12$ back into the expression for $T_{r+1}$:

$T_{13} = (-1)^{12} \binom{18}{12} 3^{36 - 3(12)}$

$T_{13} = 1 \cdot \binom{18}{12} 3^0 = \binom{18}{12} = \binom{18}{6}$

Calculating $\binom{18}{6}$:

$\binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

Simplifying the factorials:

$\binom{18}{6} = \frac{18}{6 \times 3} \times \frac{15}{5} \times \frac{16}{4} \times \frac{14}{2} \times 17 \times 13$

$\binom{18}{6} = 1 \times 3 \times 4 \times 7 \times 17 \times 13$

$\binom{18}{6} = (3 \times 4 \times 7) \times (17 \times 13)$

$\binom{18}{6} = 84 \times 221$

Given that the term independent of $x$ is equal to $(221)k$:

$221k = 84 \times 221 \implies k = 84$

Hence, the correct option is A.