Let $W_t$ be a Brownian motion, and let $F_t$ be its filtration then for $t > s$ we are asked to compute
$$\mathbb{E}\left[W_t^2|F_s\right]$$
We have $$W_t = W_s + (W_t - W_s)$$
and
$$W_t^{2} = W_s^{2} + 2W_s(W_t - W_s) + (W_t - W_s)^2$$
So
$$\mathbb{E}\left[W_t^{2}|F_s\right] = W_s^{2} + t - s$$
I don't see how
$$2\mathbb{E}\left[W_s(W_t - W_s)|F_s\right] = t - s$$
\begin{equation} \mathbb{E} \left[ \left. W_s \left( W_t - W_s \right) \right| \mathfrak{F}_s \right] = W_s \mathbb{E} \left[ W_t - W_s \right] = 0 \end{equation}
The first step uses that $W_s$ is $\mathfrak{F}_s$-measureable and that the increment $W_t - W_s$ is independent of $\mathfrak{F}_s$. Next,
\begin{equation} \mathbb{E} \left[ \left. \left( W_t - W_s \right)^2 \right| \mathfrak{F}_s \right] = \mathbb{E} \left[ \left( W_t - W_s \right)^2 \right] = \mathbb{E} \left[ W_{t - s}^2 \right] = t - s. \end{equation}
Here we used again independence in the first step. In the second one we use that the unconditional distribution of $W_t - W_s$ is the same as that of $W_{t - s}$.
