The arithmetic mean of the numbers 9, 12, 15, 18, 20, 22 is increased by 1, if 15 is replaced by:

The current sum is $9+12+15+18+20+22=96$. The number of items $n=6$.
Current mean = $\frac{96}{6}=16$.
If the mean increases by $1$, the new mean is $17$.
The new sum required is $17\times 6=102$.
Old sum excluding 15 is $96-15=81$.
Let the replacement number be $x$.
$81+x=102⟹x=21$.