A vertical tower stands inside a triangular field whose side...

Let the foot of the tower be $P$ and its height be $h$.

Since the angles of elevation of the top of the tower from the three corners of the field are equal, the horizontal distances from $P$ to the three corners must be equal. This implies that $P$ is the circumcentre of the triangular field.

The sides of the field are $6$ m, $8$ m, and $10$ m. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$, the field is a right-angled triangle.

For a right-angled triangle, the circumcentre lies at the midpoint of the hypotenuse. Thus, the circumradius $R$ is half the length of the hypotenuse:
$R = \dfrac{10}{2} = 5$ m.

From the right-angled triangle formed by the tower and the circumradius, we have:
$\tan(45^{\circ}) = \dfrac{h}{R}$

$h = R \tan(45^{\circ}) = 5 \times 1 = 5$ m.

Answer: $5$ m