A copy of Wooldridge is not in proximity. But the author is correct in stating equation no. $(4.10). $
Because it seems to obfuscate the thinking in OP, it warrants a brief casting of light on the matter.
The variance of the least square coefficient vector $\bf b$ is
\begin{align}
\mathbb{Var} [\mathbf b|\mathbf X] & =\left(\mathbf{ X^\mathsf TX}\right) ^{-1}\mathbf X^\mathsf T\mathbb E \left[\boldsymbol{\varepsilon\varepsilon}^\mathsf T|\mathbf X\right]\left(\left(\mathbf{ X^\mathsf TX}\right) ^{-1}\mathbf X^\mathsf T\right) ^\mathsf T\\ &= \sigma^2\left(\mathbf{ X^\mathsf TX}\right) ^{-1}.\tag 1
\end{align}
For estimation of $(1), $ one needs to estimate $\sigma^2.$ It is an easy routine to deduce based on sum of squared residuals
$$\mathbb E\left[\mathbf e^\mathsf T\mathbf e|\mathbf X\right] = (n - k) \sigma^2.\tag 2$$
From $(2) ,$ one gets an (unbiased) estimator of $\sigma^2$ and subsequently
$$\operatorname{Est.Var} [\mathbf b|\mathbf X] = s^2\left(\mathbf{ X^\mathsf TX}\right) ^{-1},\tag 3$$ where $$s^2 := \frac{\mathbf e^\mathsf T\mathbf e}{n-k}.$$
Now, to ensure the data in large samples is "well-behaved", one assumption taken is
\begin{align}\operatorname{plim}_{n\to \infty} \frac{\mathbf X^\mathsf T\mathbf X}n &:= \mathbf Q \\&\equiv \textrm{a positive definite matrix};\tag{A1}\end{align} this is attainable if it is assumed $$\mathbb E\left[\mathbf x_i\mathbf x_i^\mathsf T\right] = \mathbf Q, $$ which means $\sum \mathbf x_i\mathbf x_i^\mathsf T/n \overset{\mathbb P}{\to}\mathbf Q$ and hence $\operatorname{plim}\left( \frac{\mathbf X^\mathsf T\mathbf X}n\right) ^{-1} =\mathbf Q^{-1}.$
Now, the asymptotic variance, like $(1) $ would be
$$\operatorname{Asy.Var}[\mathbf b] = (\sigma^2/n) \mathbf Q^{-1}.\tag 4$$ Again, to get an estimator of $(4), $ one needs to concentrate on the term in parenthesis - $s^2$ can be used with proper evaluation of its consistency:
\begin{align}s^2 &= \frac{\boldsymbol \varepsilon^\mathsf T\mathbf M\boldsymbol\varepsilon}{n-k}\\&= \frac{\boldsymbol \varepsilon^\mathsf T\boldsymbol\varepsilon - \boldsymbol\varepsilon^\mathsf T \mathbf X\left(\mathbf{ X^\mathsf TX}\right)^{-1}\mathbf X^\mathsf T\boldsymbol \varepsilon}{n-k}\\ &= \frac{n}{n-k}\left[\frac{\boldsymbol \varepsilon^\mathsf T\boldsymbol\varepsilon}n - \left(\frac{\boldsymbol\varepsilon^\mathsf T \mathbf X}n\right)\left(\frac{\mathbf{ X^\mathsf TX}}n\right)^{-1}\left(\frac{\mathbf X^\mathsf T\boldsymbol \varepsilon}n\right)\right]; \tag 5
\end{align}
where $\mathbf M:= \mathbf I - \mathbf X\left(\mathbf{ X^\mathsf TX}\right)^{-1}\mathbf X^\mathsf T$ is the residual maker.
Using $(\mathrm A1) $ and the fact that $\operatorname{plim} \left(\frac{\boldsymbol\varepsilon^\mathsf T \mathbf X}n\right) = \mathbf O$ (since $\mathbb E[\mathbf x\varepsilon] =\mathbf O$), the second term in the bracket in $(5) $ converges to $0.$ The factor outside the bracket converges to $1.$ The only term left for inspection is $\sum \varepsilon_i^2/n.$ Now $\varepsilon_i^2$ are independent with same finite expectation $\sigma^2.$ To ensure the term's convergence (almost surely to $\sigma^2$), Markov's law of large numbers could be employed provided $\mathbb E\left[\left(\varepsilon_i^2\right)^{1+\delta}\right] <\infty$ for some $\delta \in (0, 1) $ - it would suffice then to assume for all $\varepsilon_i,$ there exist finite moments greater than $2.$ In any case, $$\operatorname{plim} s^2 = \sigma^2. \tag 6$$ This implies
$$ \operatorname{plim} s^2 \left( \frac{\mathbf X^\mathsf T\mathbf X}n\right) ^{-1}= \sigma^2\mathbf Q^{-1}. \tag{ 6.I}$$ Therefore, the appropriate estimator of $(4)$ would be $$\operatorname{Est.Asy.Var}[\mathbf b] = s^2\left(\mathbf X^\mathsf T\mathbf X\right) ^{-1}.\tag 7$$
Reference
Econometric Analysis, William H. Greene, Pearson Education, 2018.
