The following statement $\left(p\rightarrow q\right)\left[(~\sim p\rightarrow q)\rightarrow q\right]$ is
$$(p\rightarrow q)\left[\sim(\sim p \rightarrow q) \vee q\right] $$ $$(p\rightarrow q)\left[\sim(p \vee q) \vee q\right] $$ $$(p\rightarrow q)\left[\left(\sim p \land \sim q\right) \vee q\right]$$ $$(p\rightarrow q)\left[\left(\sim p \vee q\right) \land \left(\sim q \vee q\right)\right]$$ $$(p\rightarrow q)\left[\left(\sim p \vee q\right) \land \left(\sim q \vee q\right)\right]$$
$$(p\rightarrow q)\left[\left(\sim p \vee q\right) \land t\right] \text {where t denotes tautology}$$ $$(p\rightarrow q)\left[\left(\sim p \vee q\right)\right] $$ $$(p\rightarrow q)\left[p\rightarrow q\right] $$
Now from here how to solve further, is there any notion of multiplication as there is no logical connective between $\left(p\rightarrow q\right)$ and $\left[p\rightarrow q\right]$
Actual answer is tautology
As Mauro points out in the Comments, there is an operator missing. Probably the statement is supposed to be:
$$\left(p\rightarrow q\right)\color{red}\to \left[(~\sim p\rightarrow q)\rightarrow q\right]$$
With that, you can proceed exactly the same way you did, and thus end up with:
$$\left(p\rightarrow q\right)\to (p \to q)$$
$$\neg(p\rightarrow q)\lor (p \to q)$$
$$t$$