The sides of square abcd .. (photo provided)

There are in fact two methods to solve this problem which I will show below:

(1) Knowing that the area of square ABCD is 144 cm^2, we can calculate the side lengths as the area of a square is given by the length of one of its sides squared, thus:
144 = l^2
l = 12 cm

Now we can take a length, let's say AD and we know that it is 12 cm; we also know that MA is 6 cm, so MD would be 18 cm.
HD = MA = 6 cm
Triangle MDH is a right-angled triangle so we can use Pythagoras' theorem (c^2 = a^2 + b^2) to solve for length MH, thus:

MH^2 = 6^2 + 18^2
MH = sq.root(360)
= 6 sq.root(10)

Now that we know the length of one of the sides of the larger square we can square it to obtain the area:

Area = (6 sq.root(10))^2 = 360 cm^2

(2) After calculating that the side length of square ABCD is 12 cm, MD is 18 cm and HD is 6 cm, we can find the area of triangle MDH using the formula below:

Area = 0.5*bh
= 0.5*6*18
= 54 cm^2

Now looking at the diagram you can actually see that there are four of these triangles (MDH, MAW, WBT and HCT) and so if we multiply the area of one of the triangles by four we can find the area of the larger square that is not occupied by the smaller square, thus:

Area = 54*4 = 216 cm^2

Now we can add this together with the area of the smaller square to get:

Area of larger square = 216 + 144 = 360 cm^2

Thus in either case the answer would be A. Hope that helps :)