Consider the Stochastic process $X(t) = \int_o^t B(s)ds$.
We know that $\int_o^t B(s)ds = \int_o^t (t - s)dB(s)$. This implies that quadratic variation of X, $<X,X>(t) = t^3/3$.
On the other hand, we know that $|B(s,w)|$ is bounded by $M(w)$ on $[0,t]$ for almost all $w$. This implies that variation of X(t), $FV(X(t,w)) < M(w)*t$ on $[0,t]$. This implies that quadratic variation of $X(t)$ is $0$ almost everywhere. How is this possible?
The second one is correct, the quadratic variation of $X$ is $0$ almost everywhere. It looks like you computed $\langle X,X \rangle_t = \int_0^t (t-s)^2ds$ the first time to get $\frac 13 t^3$, but this doesn't work because the formula that $\langle \int_0^t h(s)dB(s),\int_0^t h(s)dB(s) \rangle = \int_0^t h(s)^2ds$ requires that $h$ depend only on $s$. Since $(t-s)$ also depends on $t$, this will not give the correct quadratic variation.