A natural number exceeds three times its square root by $4$. Find the number.

The correct option is A $16$
Let the number be $n$
$n-3\sqrt{n}=4$
$n-4=3\sqrt{n}$
Squaring both sides:
$(n-4{)}^2=9n$
$n^2-17n+16=0$
$n^2-16n-n+16=0$
$(n-16)(n-1)=0$
$\therefore n=1,16$
$n=1$ doesn't satisfy the condition, Hence, $n=16$