= Solution
For $t>0$, write $A_t=t^{-1}\int_0^tW_u\,du$. Deterministic integrals of the <Gaussian process> $W$ are Gaussian, as follows by taking $L^2$ limits of Riemann sums, so $(W,A)$ is jointly Gaussian. For $0<s\leq t$, integrating the <Brownian covariance kernel> gives
$$
\begin{aligned}
\mathbb E[W_sA_t]&=s-\frac{s^2}{2t},&
\mathbb E[A_sW_t]&=\frac s2,\\
\mathbb E[A_sA_t]&=\frac s2-\frac{s^2}{6t}.
\end{aligned}
$$
For the last equality, integrate $\int_0^t\min(u,v)\,dv=ut-u^2/2$ over $0\leq u\leq s$ and divide by $st$. Consequently
$$
\boxed{
\mathbb E[(W_s-cA_s)(W_t-cA_t)]
=s+c(c-3)\left(\frac s2-\frac{s^2}{6t}\right).}
$$
The only nonzero choice giving Brownian covariance is $\boxed{c=3}$; even the variance condition at $s=t$ requires $c(c-3)=0$.
Define $\widehat W_0=0$. Since $|A_t|\leq\sup_{u\leq t}|W_u|\to0$ almost surely as $t\downarrow0$, the transformed process is continuous at zero, as well as at positive times. It is centered Gaussian and has covariance $s\wedge t$, so part (c) proves
$$
\boxed{\widehat W_t=W_t-\frac3t\int_0^tW_u\,du,\qquad \widehat W_0=0.}
$$
\b[The transformed process is Brownian in its own natural filtration.] This is the <Brownian motion transform by three times its running average>. The filtration qualification matters: for $t>s>0$,
$$
\mathbb E[\widehat W_t-\widehat W_s\mid\mathcal F_s^W]
=\frac{3(t-s)}t(A_s-W_s),
$$
which is not identically zero. The Gaussian covariance calculation identifies the Brownian filtration generated by $\widehat W$; it does not make the transformed process a <martingale> in the larger original Brownian filtration.
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