Theorem 3.6 i) here (Pratulananda Das and Ekrem Savas: On $I$-convergence of nets in locally solid Riesz spaces, Filomat 27:1 (2013), 89–94, DOI:10.2298/FIL1301089D)
$3.6$ i) $I_τ-\lim s_α = x_0 ⇒ I_τ-\lim a.s_α = a.x_0$ for each $a ∈ \mathbb R$
proof: Let $U$ be a $τ$-neighborhood of zero. Choose $V ∈ N_{sol}$ such that $V ⊂ U$. Since $I-\lim s_α = x_0$, $\{α ∈ D : s_α − x_0 ∈ V\} ∈ F(I)$. Let $|a| ≤ 1$. Since $V$ is balanced, $s_α − x_0 ∈ V$ implies that $a (s_α − x_0) ∈ V$. Hence we have $$\{α ∈ D : s_α − x_0 ∈ V\} ⊂ \{α ∈ D : a.s_α − a.x_0 ∈ V\} ⊂ \{α ∈ D : a.s_α − a.x_0 ∈ U\}$$
and so $$\{α ∈ D : a.s_α − a.x_0 ∈ U\} ∈ F (I).$$
Now let $|a| > 1$ and as usual let $[|a|]$ be the smallest integer greater than or equal to $|a|$. There exists a $\color{blue}{W ∈ N_{sol} \text{ such that [|a|] W ⊂ V}}$. Since $I_τ-\lim s_α = x_0$, $$A = \{α ∈ D : s_α − x_0 ∈ W\} ∈ F (I)$$ . Then we have $$|a.x_0 − a.s_α| = |a| |x_0 − s_α| ≤ [|a|] |x_0 − s_α| ∈ [|a|] W ⊂ V ⊂ U$$ for each $α ∈ A$. Since the set $V$ is solid, we have $a.s_α − a.x_0 ∈ V$ and so $a.s_α − a.x_0 ∈ U$ for each $α ∈ A$. So we get $$\{α ∈ D : a.s_α − a.x_0 ∈ U\} ⊃ A$$ and so it belongs to $F(I)$. Hence $I_τ-\lim a.s_α = a.x_0$ for every $a ∈ \mathbb R$
Now the $\color{blue}{ coloured}$ part , how do we find that $W$? From the definition of $N_{sol}$ we can say we have $V_1$ s.t $V_1+V_1\subset W$ and then $V_2\in N_{sol}$ s.t. $V_2+V_2\subset V_1$ and thus $$V_2+V_2+V_2+V_2\subset W$$ and so on. going this way we can, at one stage, find that $nV\subset W$ where $n$ is even number. What happens if $|[a]|$ is odd?
Hope I could make my problem clear to you. pls help.thanks.
You know that for each $k$ there exists $V$ such that $$2^k V \subseteq W.$$ (This can be done in the way you described.)
Now you can simply take $k$ large enough to have $n\le 2^k$ and you get $$n V \subseteq 2^k V \subseteq W.$$ (Just notice that if $V\ni 0$ and $s\le t$ are positive integers, then $sV\subseteq tV$. If we apply this for $s=n$ and $t=2^k$, we get the above inclusion.)1
However, this is used to prove that $\newcommand{\Ilim}{\operatorname{I-lim}}\Ilim s_\alpha =x_0$ implies $\Ilim as_\alpha=ax_0$. And the proof given in the paper seems to be unnecessary complicated.
We know that $f \colon x\mapsto ax$ is a continuous function. (This is true for arbitrary topological vector space.)
So for any convergent net and any ideal we have $$s_\alpha \to x_0 \qquad \implies \qquad f(s_\alpha) \to f(x_0),$$ i.e., $$as_\alpha \to ax_0 \qquad \implies \qquad as_\alpha \to ax_0.$$
Similarly, to see that Theorem 3.6(ii) holds, we can just use continuity of addition as the map $V\times V\to V$, and the claim immediately follows. (Basically, I am just saying that we do not need to reprove from scratch facts which are already known, or things which can be easily derived from some well-known facts.)
1Since (based on the comments) this part seemed unclear: If $x\in sV$, this means that $x=v_1+\dots v_s$ for some $v_1,\dots,v_2\in V$. And we can now rewrite $x$ as $$x=v_1+\dots v_s+\underset{\text{$(t-s)$-times}}{\underbrace{0+\dots+0}}$$ to see that $x$ also belongs to $tV$.