Let f: R → R be a function which satisfies f(x + y) = f(x) + f(y),...

$∵ f\left(x + y\right) = f\left(x\right) + f\left(y\right)$

$∴ f\left(x\right) = kx, \forall x \in R$

Now $f\left(1\right) = 2 ⇒ k = 2$

$∴ f\left(x\right) = 2x, \forall x \in R$

$g\left(n\right) = f\left(1\right) + f\left(2\right) + \ldots \ldots . . + f\left(n - 1\right)$

$= 2\left[1 + 2 + 3 + \ldots \ldots + \left(n - 1\right)\right]$

$= 2 \cdot \dfrac{\left(n - 1\right)}{2}n = 20$

$⇒ {n}^{2} - n - 20 = 0$

$⇒ n = 5$.