As you haven't said anything about the material are you looking for the magnetic induction? H = B/(mu0) - M where M is the magnetization vector which depends on the material. Since these are the same in free space I suppose it makes no real difference, but this problem is commonly done in terms of B not H.

Anyway, you have a current loop, therefore you should have a magnetic field. The overall magnetic field will be the vector sum of field due to each line current. The magnetic field due to a line current is:

B = (mu0)I/(2(pi)r)*(sin(alpha1) + sin(alpha2))

where r is the perpendicular distance of the point (where we are finding B) to the line current, (alpha1) is the angle between the perpendicular to the line current and a line from the point where you are finding the field and the beginning of the line current, and (alpha2) is the same as angle 1 but measured from the point to the end of the line current.

Since it appears the current loop is "centered" on the origin the formula becomes:

B = (mu0)I/(2(pi)r)*sin(alpha)

where (alpha) is the angle between the perpendicular to the line current and the relevant diagonal of the rectangle.

So consider the top current element. sin(alpha) = (b/2)/(sqrt{a^2 + b^2}/2) = a/sqrt{a^2 + b^2}. Thus

B(top) = (mu0)I/(2(pi)*a/2)*b/sqrt{a^2 + b^2} = (mu0)Ib/(a*(pi)*sqrt{a^2 + b^2})

The direction of this field is given by the right hand rule and is thus down at the origin. ("Down" meaning into the plane formed by the current loop.)

Similarly the bottom current element gives the same field with the same expression, so

B(bottom) = (mu0)Ib/(a*(pi)*sqrt{a^2 + b^2})

Consider the side current elements. I get:

B(right) = (mu0)Ia/(b*(pi)*sqrt{a^2 + b^2})

B(left) = (mu0)Ia/(b*(pi)*sqrt{a^2 + b^2})

Both fields are also downward.

So the overall magnetic field will be the vector sum of these, which are all conveniently in the same direction:

B = (mu0)I/((pi)*sqrt{a^2 + b^2}) * ( 2b/a + 2a/b)

B = 2*sqrt{a^2 + b^2}*(mu0)I/(ab*(pi))

and the direction is downward.

-Dan