If $B$ is in $SPACE(n^2)$ and $A \leq_p B$ then so $A$ will be in $SPACE(n^2)$?
$A$ may or may not be in $\mathrm{SPACE}(n^2)$.

Any language $A\in\mathrm{P}$ can be reduced be reduced to any nontrivial language $B\in \mathrm{SPACE}(n^2)$ in polynomial time.  Fix your favourite strings $x_\mathrm{yes}\in B$ and $x_{\mathrm{no}}\notin B$ ("nontrivial" means that $B$ is neither $\emptyset$ nor $\Sigma^*$, so both of these strings exist). The reduction is to determine if the input is in $A$; if so, map it to $x_\mathrm{yes}$; if not, map it to $x_\mathrm{no}$.

This is true regardless of whether $A\in\mathrm{P}\cap\mathrm{SPACE}(n^2)$ or $A\in\mathrm{P}\setminus\mathrm{SPACE}(n^2)$.

To get some intuition, it might help to look at it in another way. To show that $A\in\mathrm{SPACE}(n^2)$, you'd want to show that the algorithm "Reduce to&nbsp;$B$, then decide&nbsp;$B$" runs in quadratic space. However, the reduction from $A$ to $B$ might blow up the length of the instance by any polynomial amount.  For example, It could map a instances of length&nbsp;$n$ to ones of length&nbsp;$n^2$. So, now, if you try to decide&nbsp;$A$ by first reducing to&nbsp;$B$, you'll need space $(n^2)^2 = n^4$ to solve  the instance of&nbsp;$B$. Or the reduction itself might take more than $n^2$&nbsp;space.