= Solution
For $t>1$, $B_{t^2}$ is not measurable with respect to the original Brownian filtration $\mathcal F_t$: its conditional variance is $t^2-t>0$. Thus it is not even adapted on the full half-line. It also fails the <martingale> condition before time one. With $s=1/4$, $t=3/4$,
$$
\mathbb E[B_{t^2}\mid\mathcal F_s]=B_s\ne B_{s^2}=M_s.
$$
The appropriate filtration is
$$
\boxed{\mathcal G_t=\mathcal F_{t^2}}.
$$
Brownian <independent increments> make $M_t=B_{t^2}$ a continuous $\mathcal G_t$-martingale, with <quadratic variation> $[M]_t=t^2$. Define the <stochastic integral>
$$
W_t=\int_0^t\frac{1}{\sqrt{2s}}\,dM_s.
$$
The value of the integrand at zero is immaterial; it is square-integrable against the bracket because $\int_0^t(2s)^{-1}\,d(s^2)=t$. Consequently $[W]_t=t$, and the <Lévy characterization of Brownian motion> makes $W$ a $\mathcal G_t$-Brownian motion. Associativity of <stochastic integrals> gives the <Brownian motion under a quadratic time change> representation
$$
\boxed{M_t=\int_0^t\sqrt{2s}\,dW_s,\qquad f(s)=\sqrt{2s}}.
$$
The function $f$ is continuous at zero. One can first verify the inverse-integrand identity on $[\varepsilon,t]$ and then let $\varepsilon\downarrow0$ using the <Itô isometry>, since $\mathbb EM_\varepsilon^2=\varepsilon^2$.
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