Halting problem under spatial restrictions
Yes (for Turing machines)

$A(n,B,d)$ can decide whether $B(n,d)$ will halt.

Note the fact that $B$'s writable memory is limited to $n$ units. This means that the number of possible states of $B$ is just $2^n$ (each possible memory configuration that $B$ may assume). Note also that $B$ halts iff it never goes back to a previously visited state, otherwise it would loop. This means that, if $B$ halts, it cannot do more than $2^n-1$ state transitions (as each state is visited only once).

Thus we can exploit the $2 \cdot n + c$ writable memory units of $A$ to build a machine which simulates B and counts the number of states it visits. We can do that by keeping the memory of $B$ in the first $n$ units, and a state counter $\gamma$ in some other $n+1$ units. We increment $\gamma$ at each state transition of $B$.

If $B$ halts, sooner or later the simulation of $B$ will reach a "HALT" instruction, and then we know $B$ halts. If $B$ does not halt, $\gamma$ will increase forever: as soon as $\gamma = 2^n$, we know that $B$ does not halt.