For set $A$

$\left(x^2-10x+9\right)\le 0\Rightarrow \left(x-1\right)\left(x-9\right)\le 0$

$\Rightarrow x\in \left[1,9\right]$

i.e. $A=\left\{1,2,3,4,5,6,7,8,9\right\}$

Given $B=\left\{n^2:n\in N\right\}$

Now $f\left(x\right)\le {\left(x-3\right)}^2+1$

For $x=1: f\left(1\right)\le 5\Rightarrow B\in \left\{1^2,2^2\right\}$

For $x=2:f\left(2\right)\le 2\Rightarrow B\in \left\{1^2\right\}$

For $x=3:f\left(3\right)\le 1\Rightarrow B\in \left\{1^2\right\}$

For $x=4:f\left(4\right)\le 2\Rightarrow B\in \left\{1^2\right\}$

For $x=5:f\left(5\right)\le 5\Rightarrow B\in \left\{1^2,2^2\right\}$

For $x=6:f\left(6\right)\le 10\Rightarrow B\in \left\{1^2,2^2,3^2\right\}$

For $x=7: f\left(7\right)\le 17\Rightarrow B\in \left\{1^2,2^2,3^2,4^2\right\}$

For $x=8:f\left(8\right)\le 26\Rightarrow B\in \left\{1^2,2^2,3^2,4^2,5^2\right\}$

For $x=9:f\left(9\right)\le 37\Rightarrow B\in \left\{1^2,2^2,3^2,4^2,5^2,6^2\right\}$

Total number of such function

$=2\times 1\times 1\times 1\times 2\times 3\times 4\times 5\times 6=2\left(720\right)=1440$