Situated in a horizontal plane, a circular loop of radius 4.0 cm...

The magnetic field at a distance $x$ on the axis of a circular loop of radius $R$ carrying current $I$ is given by:

$B = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$

Given values are:
$I = 5.0 \text{ A}$
$R = 4.0 \text{ cm} = 4.0 \times 10^{-2} \text{ m}$
$x = 3.0 \text{ cm} = 3.0 \times 10^{-2} \text{ m}$
$\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$

Calculating the denominator term:
$R^2 + x^2 = (4.0 \times 10^{-2})^2 + (3.0 \times 10^{-2})^2 = 25 \times 10^{-4} \text{ m}^2$
$(R^2 + x^2)^{3/2} = (25 \times 10^{-4})^{3/2} = 125 \times 10^{-6} \text{ m}^3$

Substituting the values into the formula:
$B = \dfrac{4\pi \times 10^{-7} \times 5.0 \times (16 \times 10^{-4})}{2 \times 125 \times 10^{-6}}$

$B = \dfrac{320\pi \times 10^{-11}}{250 \times 10^{-6}}$

$B = 1.28\pi \times 10^{-5} \text{ T}$

Using $\pi \approx 3.14$:
$B \approx 4.0 \times 10^{-5} \text{ T}$

Since the current is in the clockwise direction when viewed from above, applying the right-hand thumb rule (curling the fingers of the right hand in the direction of current makes the thumb point downwards) gives the direction of the magnetic field as downwards.