Find a three-digits number $\overline{abc}$ such that $\overline{abc}^2=(a+b+c)^5.$
It is easy to see that $$ (a+b+c)^5 \leq 999^2 \implies a+b+c< \sqrt[5]{999^2}\leq 15 $$ and $$ (100 a+10b+c)^2<15^5 \implies 100 a+10b+c>\sqrt{15^5} \leq 871. $$ Also $$ (100 a+10b+c)^2 =(a+b+c) \mod 2 $$ implies $a+b=0 \mod 2$. Similarly $$ (100 a+10b+c)^2 = c^2 = (a+b+c) \mod 5. $$ But it not enough to find the solution $243$. No more ideas.
$(a+b+c)$ must be a square, so it can only be $1,4,9,16$ or $25$.
By your inequalities we have $a+b+c=1,4$ or $9$.
notice that $(100a+10b+c)^2\geq 100^2$.On the other hand $1^5$ and $4^5$ are too small.
We conclude that $a+b+c=9$.
So now we must have $(100a+10b+c)=\sqrt{9^5}=3^5=243$
You should be able to quickly see that $(a+b+c)$ must be a square. Since $999<1024$, we know that $(a+b+c)<16$ so we have ${a+b+c}\in\{1,4,9\}$, and thus $\sqrt{(a+b+c)^5}\in\{1,32,243\}$, the last of which produces a valid solution. Since you form $\overline{abc}$ as a number I'll assume $a$ cannot be zero.
Let $$x = \sqrt{a+b+c} = \sqrt[5]{\overline{abc}},\quad a\in \{1, 2\dots9\}, \quad b,c\in\{0,1\dots9\},$$ then $$\begin{cases} \sqrt{1} <= x < \sqrt{27}\\ \sqrt[5]{100} < x < \sqrt[5]{999}\\ x\in\mathbb N, \end{cases}$$ $$\begin{cases}2.5 < x < 3.99\\ x\in\mathbb N, \end{cases}$$ $$x=3,$$ $$\begin{cases} \overline{abc} = x^5 = 243\\ a+b+c = x^2 = 9, \end{cases}$$ $$\boxed{a=2,\quad b=4,\quad b=3}.$$