Write $\Delta_iM=M_{t_i}-M_{t_{i-1}}$ and $Q=\sum_i(\Delta_iM)^2$. The identity
$$
Q=M_t^2-M_0^2-2\sum_iM_{t_{i-1}}\Delta_iM
$$
shows, using the <martingale-difference orthogonality>, that
$$
\begin{aligned}
\mathbb E[Q^2]
&\leq2\mathbb E[(M_t^2-M_0^2)^2]
 +8\mathbb E\left[\left(\sum_iM_{t_{i-1}}\Delta_iM\right)^2\right]\\
&\leq2C^4+8C^2\mathbb E[Q].
\end{aligned}
$$
Also $\mathbb E[Q]=\mathbb E[M_t^2]-\mathbb E[M_0^2]\leq C^2$. Hence $\mathbb E[Q^2]\leq10C^4$, which in particular proves the requested bound
$$
\boxed{\mathbb E[Q^2]\leq48C^4.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 2 / 1 / Solution

If $(M_n)$ is a square-integrable <martingale>, then its increments are orthogonal in $L^2$: for $i<j$,
$$
\mathbb E[(M_i-M_{i-1})(M_j-M_{j-1})]=0.
$$
More generally, multiplying the later increment by any square-integrable quantity measurable before it still gives expectation zero whenever the product is integrable.

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 Incoming links: Martingale-difference orthogonality