Find the value of the problem

$\displaystyle \sum_{k=1}^{20}f(k)=f(1)+f(2)+f(3)+\cdots \cdots+f(19)+f(20)$

The second condition says: $f(m)<f(n)$ if $m<n$ .

We know $1<2$

Thus, $f(1)<f(2)$ . So, $f(1)=1$ [as it is a set of 'Natural numbers']

The first condition says: $f(mn)=f(m)f(n)$

So, $f(4)=f(2\cdot 2)=f(2)f(2)=2\times 2$ [according to the third condition]

Hence, $f(4)=4$

As we know $2<3<4$ so, with the help of second condition we get:

Thus, $f(2)<f(3)<f(4)$ which is $2<f(3)<4$

Hence, $f(3)=3$

Similarly, we can find with the help of the conditions: $f(5)=5; f(6)=6; f(7)=7; \cdots \cdots ;f(19)=19; f(20)=20$ which means $f(k)=k$

Therefore, $\displaystyle \sum_{k=1}^{20} f(k)=\displaystyle \sum^{20}_{k=1} k=1+2+3+\cdots\cdots+19+20=\boxed{210}$