= Solution
\b[The overshoot hypothesis as printed cannot hold for every $R>0$.] For $R\leq1$, $T_R=0$ and $M_{T_R}=1$; choosing $R<1/\tau$ contradicts $M_{T_R}\leq\tau R$. The intended and sufficient assumption is the overshoot bound for $R\geq1$, on the event $\{T_R<\infty\}$. We use those thresholds below; when $R\leq1$, the requested stopped-time bound is trivial because $T\wedge T_R=0$.
Fix $R\geq1$ and set $S=T\wedge T_R$. By part (a), the stopped <martingale> satisfies $0\leq M_{S\wedge n}\leq\tau R$: before stopping it is below $R$, at $T$ it is zero, and at $T_R$ the assumed overshoot bound applies. Thus the square of this stopped process is integrable. For $\Delta M_n=M_{n+1}-M_n$,
$$
X_{n+1}-X_n
=\mathbf1_{\{S>n\}}\left(2M_n\Delta M_n+(\Delta M_n)^2-\sigma^2\right).
$$
The event $\{S>n\}$ is $\mathcal F_n$-measurable and contained in $\{T>n\}$. <Conditional expectation> of the linear term is zero, while the <variance> assumption makes the remaining <conditional expectation> nonnegative. The stopped increments here are bounded by the stopping and overshoot bounds, so all <conditional expectations> are legitimate. Consequently
$$
\boxed{(X_n)_{n\geq0}\text{ is a submartingale}.}
$$
Since $X_0=1$, we have $\sigma^2\mathbb E(S\wedge n)\leq\mathbb E M_{S\wedge n}^2-1$. Bounded-time <optional stopping theorem> also gives $\mathbb E M_{S\wedge n}=1$, so
$$
\mathbb E M_{S\wedge n}^2\leq\tau R\,\mathbb E M_{S\wedge n}=\tau R.
$$
Apply the <monotone convergence theorem> to $S\wedge n$. This proves the stronger estimate $(\tau R-1)/\sigma^2$ and hence the requested <absorption-time bound from conditional variance and overshoot>:
$$
\boxed{\mathbb E(T\wedge T_R)\leq\frac{\tau R-1}{\sigma^2}\leq\frac{\tau^2R}{\sigma^2}\qquad(R\geq1).}
$$
In particular this stopping time is finite almost surely. For $0<R\leq1$, $S=0$ and $X_n=1$ is a constant <submartingale>, while the stopping-time <expectation> is zero.
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