= Solution
Let $\mathcal F_s=\sigma(W_r:0\leq r\leq s)$. For $0\leq s<t$, the increment $W_t-W_s$ is independent of $\mathcal F_s$ and distributed as $\sqrt{t-s}Z$. Apply part (a) to the translated function $u\mapsto f(W_s+u)$, conditionally on $\mathcal F_s$:
$$
\begin{aligned}
\mathbb E[f(W_t)\mid\mathcal F_s]
&=f(W_s)
+\frac12\int_0^{t-s}
\mathbb E[f''(W_s+\sqrt rZ)\mid\mathcal F_s]\,dr\\
&=f(W_s)
+\frac12\mathbb E\left[
\int_s^t f''(W_r)\,dr\ \middle|\ \mathcal F_s
\right].
\end{aligned}
$$
Therefore
$$
\begin{aligned}
\mathbb E[M_t\mid\mathcal F_s]
&=\mathbb E[f(W_t)\mid\mathcal F_s]
-\frac12\int_0^sf''(W_r)\,dr\\
&\quad-\frac12\mathbb E\left[
\int_s^tf''(W_r)\,dr\ \middle|\ \mathcal F_s
\right]\\
&=f(W_s)-\frac12\int_0^sf''(W_r)\,dr
=M_s.
\end{aligned}
$$
The assumed growth conditions supply integrability, so $(M_t)$ is the <Brownian compensator martingale>.
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