Solution
= Solution
{c}
The given <Doob L2 maximal inequality> gives
$$
\mathbb E[(M_t^*)^2]\leq4\mathbb E|M_t|^2<4\,2^{-2n}=2^{-2(n-1)}.
$$
Apply <Markov's inequality> to the nonnegative variable $(M_t^*)^2$, at the threshold printed in the question:
$$
\boxed{\mathbb P\bigl((M_t^*)^2\geq2^{-(n-1)}\bigr)
\leq\frac{2^{-2(n-1)}}{2^{-(n-1)}}=2^{-(n-1)}.}
$$
The corresponding threshold for $M_t^*$ itself is $2^{-(n-1)/2}$.