Question:

Step 1: To find the term without $x$ in the expansion of $\left(x^{2/3}+\dfrac{\alpha}{x^3}\right)^{22}$, we need to use the Binomial Theorem, which states that for any non-negative integer $n$, the expansion of $(a + b)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^{k}$. Here, $a = x^{2/3}$ and $b = \dfrac{\alpha}{x^3}$. Step 2: The general term in the expansion of $\left(x^{2/3}+\dfrac{\alpha}{x^3}\right)^{22}$ is given by $\binom{22}{k}(x^{2/3})^{22-k}(\dfrac{\alpha}{x^3})^{k}$. For the term without $x$, the powers of $x$ must cancel out, so we need to find the value of $k$ for which this occurs. The power of $x$ in the general term is $\dfrac{2(22-k)}{3} - 3k$. Step 3: Setting the power of $x$ equal to zero, we get $\dfrac{2(22-k)}{3} - 3k = 0$. Solving for $k$, we have $\dfrac{44 - 2k}{3} = 3k$, which simplifies to $44 - 2k = 9k$, and then $44 = 11k$, so $k = 4$. This means the term without $x$ is $\binom{22}{4}(x^{2/3})^{18}(\dfrac{\alpha}{x^3})^{4}$. Step 4: Substituting $k = 4$ into the general term, we get $\binom{22}{4}(\dfrac{\alpha}{x^3})^{4}(x^{2/3})^{18} = \binom{22}{4}\alpha^{4}x^{-12}x^{12} = \binom{22}{4}\alpha^{4}$. Since $\binom{22}{4} = \dfrac{22!}{4!(22-4)!} = \dfrac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} = 7315$, we have $7315\alpha^{4} = 7315$. Step 5: Dividing both sides of the equation $7315\alpha^{4} = 7315$ by 7315, we obtain $\alpha^{4} = 1$. Taking the fourth root of both sides, we get $\alpha = \pm 1$. Since we are asked to find $|\alpha|$, the absolute value of $\alpha$, we conclude that $|\alpha| = 1$. Step 6: Therefore, the final answer is $|\alpha| = 1$, which corresponds to option 1. The final answer is $\boxed{1}$.