Comparing areas

First we need to find the length of each rectangle. Using the formula of perimeters, the length of rectangle $A$ in units is:

$2L_A + 2W_A = 32 \implies (2\times 6)+ 2L = 32 \implies L_A= 10$ units.

Similarly, for rectangle $B$ , we can find its length:

$2L_B + 2W_B = 32 \implies (2\times 4)+ 2L = 32 \implies L_B= 12$ units.

Thus, the Area of Rectangle $A$ is:

$L_A\times W_A= 10\times 6= 60$ square units.

and the Area of Rectangle $B$ is:

$L_B\times W_B= 12\times 4= 48$ square units.

Comparing the two areas yields $\boxed{A}$ is the greater area.