That answer, obviously, corresponds to a different exercise, which can be formulated as follows: What is ${\rm Cov}(W(s),W(t))$, $0 \leq s \leq t$, when $W(t)=t B(1/t)$ and $W(0)=0$, where $B$ is a standard BM. [The point in defining $W(0)=0$ is that $1/0$ is not defined; further, note that this definition agrees with the variance of $W(t)$ as $t \downarrow 0$.]
General remark: by symmetry, it suffices to calculate covariance functions for $s \leq t$.
EDIT: The purpose of this exercise is to show that by a simple transformation, namely $W(t)=t B(1/t)$, one BM can be transformed into another. Note that the processes $W:=\{W(t):t \geq 0\}$ and $B:=\{B(t):t \geq 0\}$ are equal only at two points, $t=0$ and $t=1$, yet they have exactly the same law (both are standard BM). [Note that the law of a mean zero Gaussian process is completely determined by its covariance function; the fact that $W$ is a mean zero Gaussian process follows straightforwardly from the fact that $B$ is.]
So, you want to show that $W$ has the same covariance function as $B$. Recall that, by symmetry, it suffices to calculate the covariance function for $0 \leq s \leq t$. In particular, letting $s=0$, you want to show that ${\rm Cov}(W(0),W(t)) = \min\{0,t\}$, for any $t \geq 0$. This is obvious.