Answer
The minimum value of $g(a)$ is $E\left[(X-\mu)^2\right]$, the $\underline{\text{variance}}$ of $X$.
Work Step by Step
$\begin{align*}
g(a) &= E\left[(X-a)^2\right] \\
&= E\left[\biggl((X-\mu) + (\mu -a)\biggr)^2\right] \\
&= E\left[(X-\mu)^2-2(X-\mu)(\mu-a) + (\mu-a)^2\right] \\
&= E\left[(X-\mu)^2\right] -E[2(X-\mu)(\mu-a)] + E\left[(\mu-a)^2\right] \quad [\ \text{Corollary 3.5.1, p. 148}\ ]\\
&= E\left[(X-\mu)^2\right] -2(\mu-a)E[(X-\mu)] + (\mu-a)^2 \quad [\ \text{Corollary 3.5.1, p. 148}\ ] \\
&= E\left[(X-\mu)^2\right] -2(\mu-a)(E[X]-\mu) + (\mu-a)^2 \quad [\ \text{Corollary 3.5.1, p. 148}\ ] \\
&= E\left[(X-\mu)^2\right] -2(\mu-a)(\mu-\mu) + (\mu-a)^2 \quad [\ \text{since}\ E[X]=\mu\ ] \\
&= E\left[(X-\mu)^2\right] -0 + (\mu-a)^2 \\
g(a) &= E\left[(X-\mu)^2\right] + (\mu-a)^2 \quad\quad\color{blue}{\text{(*** Eqn. 1 ***)}}
\end{align*}$
Note from Equation 1 that $g(a)$ is minimum if $\mu=a$ (so that the second term on the right-hand side is zero). Thus, the minimum value of $g(a)$ is $g(a) = E\left[(X-\mu)^2\right]$, this being the $\underline{\text{variance}}$ of $X$ (see Definition 3.6.1, p. 154).
