The square of a number is $48$ more than $22$ times of itself. Find the number.

Let the required number be $x$.
According to the problem statement:

$$x^2=22x+48$$

Rearranging into a standard quadratic equation format ($a{x}^2+bx+c=0$):

$$x^2-22x-48=0$$

Factorize the quadratic expression by splitting the middle term:

$$x^2-24x+2x-48=0$$

$$x(x-24)+2(x-24)=0$$

$$(x-24)(x+2)=0$$

This yields two possible values for $x$:

$$x=24\text{or}x=-2$$

Since the options only consist of positive values, the number is $24$.